Explanation:
Okay, so I need to find a vector v such that when I take the cross product of the vector (1, 2, -5) with v, I get the vector (90, 30, 30). And among all possible such vectors v, I need the one with the smallest magnitude. Hmm, cross products... Let me recall how cross products work.

First, the cross product of two vectors in three dimensions results in a vector that's perpendicular to both of the original vectors. So, if I have vectors a and b, then a × b is perpendicular to both a and b. That means that the cross product here, which is (1, 2, -5) × v = (90, 30, 30), must be perpendicular to both (1, 2, -5) and v. Wait, but maybe that's not directly helpful here. Let me think.

Alternatively, maybe I can set up equations. Let me denote vector v as (x, y, z). Then, the cross product of (1, 2, -5) and (x, y, z) is given by the determinant of a matrix with the standard unit vectors i, j, k in the first row, the components of the first vector in the second row, and the components of the second vector in the third row. Let me write that out.

So, cross product:

|i   j   k |
|1   2  -5|
|x   y   z|

Which is i*(2*z - (-5)*y) - j*(1*z - (-5)*x) + k*(1*y - 2*x)

Simplifying each component:

First component (i): 2z + 5y

Second component (j): -(z + 5x) → -z -5x

Third component (k): y - 2x

So, the cross product is (2z + 5y, -z -5x, y - 2x). And this is equal to (90, 30, 30). Therefore, we can set up the following system of equations:

1) 2z + 5y = 90

2) -z -5x = 30

3) y - 2x = 30

So, three equations with three variables x, y, z. Let me write them again:

Equation 1: 5y + 2z = 90

Equation 2: -5x - z = 30

Equation 3: -2x + y = 30

So, I need to solve this system for x, y, z. Let me try solving equations 2 and 3 first, since equation 3 has only x and y, and equation 2 has x and z. Maybe I can express variables in terms of others.

From equation 3: y = 2x + 30. Let me write that down.

Equation 3: y = 2x + 30.

Then, substitute this into equation 1. Equation 1 is 5y + 2z = 90. So replacing y:

5*(2x + 30) + 2z = 90

Compute that:

10x + 150 + 2z = 90

Subtract 150 from both sides:

10x + 2z = 90 - 150 = -60

Divide both sides by 2:

5x + z = -30

So, equation 1 becomes 5x + z = -30. But equation 2 is -5x - z = 30. Wait, equation 2 is -5x - z = 30. Let me check:

Original equation 2: -z -5x = 30, which is same as -5x - z = 30. So, if I have equation 1 transformed as 5x + z = -30, which is the same as -5x - z = 30. Wait, that's exactly equation 2. So equations 1 and 2 are dependent? So, that means that equations 1 and 2 are not independent. Therefore, we have only two equations here, but three variables. Therefore, there are infinitely many solutions, which matches the problem statement that there are infinitely many vectors v. So, we need to find the vector v with the smallest magnitude.

So, since we have two equations and three variables, we can express two variables in terms of the third. Let me see. From equation 3, we have y = 2x + 30. From equation 2, which is equivalent to equation 1 transformed, we have 5x + z = -30. Therefore, z = -30 -5x. Therefore, we can express y and z in terms of x. Therefore, vector v is (x, y, z) = (x, 2x + 30, -5x -30). Therefore, v can be written as:

v = (x, 2x + 30, -5x -30)

So, now, we need to find the value of x that minimizes the magnitude of v. The magnitude squared of v is x² + (2x + 30)² + (-5x -30)². Since the square root is a monotonic function, minimizing the magnitude is equivalent to minimizing the magnitude squared. So, let me compute that.

Let me compute the magnitude squared:

||v||² = x² + (2x + 30)² + (-5x -30)²

Let me expand each term:

First term: x²

Second term: (2x + 30)² = 4x² + 120x + 900

Third term: (-5x -30)² = (5x + 30)² = 25x² + 300x + 900

So, adding all three terms:

x² + 4x² + 120x + 900 + 25x² + 300x + 900

Combine like terms:

x² + 4x² + 25x² = 30x²

120x + 300x = 420x

900 + 900 = 1800

Therefore, ||v||² = 30x² + 420x + 1800

So, we need to find the value of x that minimizes this quadratic function. Since it's a quadratic in x with a positive coefficient on x², the minimum occurs at the vertex. The x-coordinate of the vertex is at -b/(2a). Here, a = 30, b = 420. Therefore, x = -420/(2*30) = -420/60 = -7.

So, x = -7. Therefore, substituting back into the expressions for y and z:

y = 2x + 30 = 2*(-7) + 30 = -14 + 30 = 16

z = -5x -30 = -5*(-7) -30 = 35 - 30 = 5

Therefore, the vector v is (-7, 16, 5). Let me check if this is correct.

First, let me verify that this vector satisfies the original cross product equation.

Compute (1, 2, -5) × (-7, 16, 5):

Using the cross product formula:

i*(2*5 - (-5)*16) - j*(1*5 - (-5)*(-7)) + k*(1*16 - 2*(-7))

Compute each component:

First component (i): 2*5 - (-5)*16 = 10 + 80 = 90

Second component (j): -(1*5 - (-5)*(-7)) = -(5 - 35) = -(-30) = 30

Third component (k): 1*16 - 2*(-7) = 16 + 14 = 30

So, the cross product is (90, 30, 30), which matches the given vector. Great, so that checks out.

Now, let me check the magnitude of this vector. The magnitude squared is 30*(-7)^2 + 420*(-7) + 1800. Wait, but maybe I can compute it directly:

v = (-7, 16, 5). So, magnitude squared is (-7)^2 + 16^2 + 5^2 = 49 + 256 + 25 = 49 + 256 is 305, plus 25 is 330. So, magnitude is sqrt(330). Let me check if that's indeed the minimum.

Alternatively, since we found the minimum of the quadratic function, which was 30x² + 420x + 1800. Let me compute the value at x = -7:

30*(-7)^2 + 420*(-7) + 1800 = 30*49 - 2940 + 1800 = 1470 - 2940 + 1800 = (1470 + 1800) - 2940 = 3270 - 2940 = 330. Which matches. So, the minimum magnitude squared is 330, so the magnitude is sqrt(330). Therefore, this is the vector with the smallest magnitude.

But let me just think again if there's another way to approach this problem, maybe using vector projections or something else, to confirm.

Another approach: The cross product a × v = b. Then, we can think of this as a vector equation. The cross product a × v is a vector perpendicular to both a and v. However, since we have a specific b, we can think of this as solving for v. But since cross product is not invertible in general, we have infinitely many solutions, as we saw.

But perhaps we can use the formula for solving cross product equations. Let me recall that if a × v = b, then v can be expressed as v = (a × b)/||a||² * a + something? Wait, maybe not. Let me think.

Alternatively, since a × v = b, then v must be in the plane perpendicular to b, because the cross product is perpendicular to both a and v. Wait, no. Wait, the cross product a × v is perpendicular to both a and v. Therefore, the vector b must be perpendicular to a. Wait, is that true?

Wait, if a × v = b, then b is perpendicular to a, because cross product is perpendicular to both vectors. Therefore, if a × v = b, then b is perpendicular to a. Let me check if that's the case here. The given b is (90, 30, 30). Let me check if it's perpendicular to a = (1, 2, -5). The dot product is 1*90 + 2*30 + (-5)*30 = 90 + 60 - 150 = 0. Yes! So, the dot product is zero, so they are perpendicular. Therefore, that's consistent. So, that's a good check.

Therefore, since b is perpendicular to a, then the equation a × v = b has solutions. Now, how to find v?

Another approach: Let me recall that if a × v = b, then v can be written as v = (a × b)/||a||² * a + c, where c is any vector perpendicular to a. Wait, maybe? Let me think.

Alternatively, since a × v = b, then v can be expressed as v = (a × b)/||a||² * a + k, where k is a vector such that a · k = 0. Wait, maybe not. Let me think again.

Let me consider that if I have a × v = b, then v can be written as v = (a × b)/||a||² * a + t, where t is a vector such that a · t = 0. Wait, let me check.

Let me suppose that v = (a × b)/||a||² * a + t, where t is perpendicular to a. Then, compute a × v:

a × v = a × [ (a × b)/||a||² * a + t ] = (a × (a × b))/||a||² * a + a × t

But since t is perpendicular to a, then a × t is a vector perpendicular to a. However, the first term is (a × (a × b))/||a||² * a. Let me compute a × (a × b). Using the vector triple product formula: a × (a × b) = a(a · b) - b(a · a). But since a · b = 0, as we saw earlier, this simplifies to -b||a||². Therefore, substituting back:

a × v = [ -b||a||² / ||a||² ] * a + a × t = -b a + a × t

But we want a × v = b. Therefore, -b a + a × t = b. Therefore, a × t = b + b a. Wait, this seems complicated. Maybe this approach is not the best.

Alternatively, since we have a × v = b, and we know that v can be expressed as v = (a × b)/||a||² * a + c, where c is any vector perpendicular to a. Wait, let me check.

Let me suppose that v = (a × b)/||a||² * a + c, where c is perpendicular to a. Then, compute a × v:

a × v = a × [ (a × b)/||a||² * a + c ] = (a × (a × b))/||a||² * a + a × c

Again, using the triple product formula, a × (a × b) = a(a · b) - b(a · a). Since a · b = 0, this becomes -b||a||². Therefore:

a × v = [ -b||a||² / ||a||² ] * a + a × c = -b a + a × c

But we want this to be equal to b. Therefore:

-b a + a × c = b ⇒ a × c = b + b a

Wait, this seems messy. Maybe this approach isn't helpful. Let me try plugging in numbers.

Alternatively, since we have already solved the system of equations and found the vector v = (-7, 16, 5), which gives the minimal magnitude, maybe that's sufficient. However, let me think if there's another way to find the minimal magnitude.

Alternatively, since we have v expressed in terms of x as (x, 2x + 30, -5x -30), then the magnitude squared is 30x² + 420x + 1800. To find the minimum, we can take derivative with respect to x and set to zero. But since it's a quadratic, we already know the vertex is at x = -7. So, that's consistent.

Alternatively, maybe we can think geometrically. The set of all vectors v satisfying a × v = b is a line in 3D space. The minimal magnitude vector on that line is the one closest to the origin. Therefore, we can find the vector v on the line that is closest to the origin, which is the projection of the origin onto the line. But how to compute that?

Alternatively, since we have parametric equations for the line. Wait, the solutions for v are given by v = (x, 2x + 30, -5x -30). So, this is a line in parametric form with parameter x. So, we can write this as:

v = x*(1, 2, -5) + (0, 30, -30). Wait, let me check:

If x is multiplied by (1, 2, -5), then adding (0, 30, -30) gives (x, 2x + 30, -5x -30). Yes, that's correct. Therefore, the line is parametrized as v = x*(1, 2, -5) + (0, 30, -30). Therefore, the direction vector of the line is (1, 2, -5), and a point on the line is (0, 30, -30). Therefore, to find the vector on this line closest to the origin, we can use the projection formula.

The closest point on the line to the origin is the projection of the origin onto the line. Wait, but the line is given by v = t*(1, 2, -5) + (0, 30, -30), where t is a scalar parameter. Wait, but in our case, the parameter was x, but here I can use t as the parameter. So, the line is parametrized as:

v(t) = t*(1, 2, -5) + (0, 30, -30)

So, to find the value of t that minimizes ||v(t)||². Which is exactly what we did earlier, by substituting into the magnitude squared and finding the minimum. However, maybe we can use the projection formula here.

The formula for the projection of a vector onto a line. Let me recall that if we have a line L through point p with direction vector d, then the closest point on L to a point q is p + proj_d(q - p). Wait, but here, the line is through the point (0, 30, -30) with direction vector (1, 2, -5). The origin is the point q = (0, 0, 0). Therefore, the vector from p to q is q - p = (0, 0, 0) - (0, 30, -30) = (0, -30, 30). Then, the projection of this vector onto the direction vector d = (1, 2, -5) is:

proj_d(q - p) = [(q - p) · d / ||d||²] * d

Therefore, the closest point on the line to the origin is p + proj_d(q - p). Let me compute this.

First, compute (q - p) · d:

(0, -30, 30) · (1, 2, -5) = 0*1 + (-30)*2 + 30*(-5) = 0 - 60 - 150 = -210

Then, ||d||² = 1² + 2² + (-5)² = 1 + 4 + 25 = 30

Therefore, proj_d(q - p) = (-210 / 30) * d = -7 * (1, 2, -5) = (-7, -14, 35)

Therefore, the closest point on the line to the origin is p + proj_d(q - p) = (0, 30, -30) + (-7, -14, 35) = (0 -7, 30 -14, -30 +35) = (-7, 16, 5). Which is exactly the vector we found earlier! So, that confirms that the minimal magnitude vector is (-7, 16, 5). Therefore, this method also gives the same result. So, that's reassuring.

Therefore, regardless of the method, we get the same vector. Therefore, I can be confident that this is the correct answer.

But just to recap, the steps were:

1. Set up the cross product equation, leading to three equations.

2. Realize that there are infinitely many solutions, parameterized by x.

3. Express the vector v in terms of x.

4. Compute the magnitude squared as a function of x, find its minimum by completing the square or using vertex formula.

5. Alternatively, use projection formula to find the closest point on the line to the origin.

Either way, we arrive at the vector (-7, 16, 5). Let me just check once more the magnitude squared:

(-7)^2 + 16^2 + 5^2 = 49 + 256 + 25 = 49 + 256 is 305, plus 25 is 330. So sqrt(330) is approximately 18.165... Is there a possibility that there's a smaller magnitude? Well, since we found the minimum of the quadratic function, and the quadratic is convex, there can't be a smaller value. Therefore, this must be the minimal.

Alternatively, if I think about the cross product equation, the magnitude of the cross product is equal to ||a|| ||v|| sin(theta), where theta is the angle between a and v. Therefore, ||a × v|| = ||a|| ||v|| sin(theta). Therefore, given that ||a × v|| is fixed (which is ||(90, 30, 30)||), then ||v|| is minimized when sin(theta) is maximized, i.e., when theta = 90 degrees, so that sin(theta) = 1. Therefore, the minimal ||v|| occurs when v is perpendicular to a. Wait, but is that the case?

Wait, if theta is the angle between a and v, then ||a × v|| = ||a|| ||v|| sin(theta). Therefore, if we fix ||a × v||, then ||v|| is minimized when sin(theta) is maximized. Since sin(theta) ≤ 1, the maximum value is 1, which occurs when theta = 90 degrees. Therefore, the minimal ||v|| occurs when v is perpendicular to a. Therefore, if we can find such a vector v that is perpendicular to a and satisfies a × v = b, then that would be the minimal one.

But wait, in our case, we found that v = (-7, 16, 5). Let me check if this vector is perpendicular to a = (1, 2, -5). Compute their dot product:

(-7)(1) + 16*2 + 5*(-5) = -7 + 32 -25 = 0. Yes! So, the dot product is zero. Therefore, v is indeed perpendicular to a. Therefore, this is consistent with the above reasoning. Therefore, this is the minimal ||v||.

Therefore, this gives another way to think about the problem: find a vector v perpendicular to a such that a × v = b. Since when v is perpendicular to a, sin(theta) = 1, so ||v|| is minimized. Therefore, solving for such a v would give the minimal magnitude.

But how do we find such a v? Well, since v is perpendicular to a, then v lies in the plane perpendicular to a. Also, since a × v = b, which is perpendicular to a, as we saw earlier. Therefore, the vector b lies in the plane perpendicular to a, and v is also in that plane. Therefore, we can think of this as a problem in 2D: find v in the plane perpendicular to a such that a × v = b. But since we are in 3D, but the cross product in that plane would be equivalent to a 2D cross product.

Alternatively, since v is perpendicular to a, we can write v as a vector in the plane perpendicular to a. Let me try to express this.

Let me denote that since v is perpendicular to a, then v can be written as v = c × a, where c is some vector? Wait, no. Wait, if v is perpendicular to a, then v is in the plane perpendicular to a. However, the cross product a × v is a vector perpendicular to both a and v, which is in the direction of b. Wait, but since b is also perpendicular to a, then b is in the plane perpendicular to a. Therefore, the cross product a × v = b implies that v is in the plane perpendicular to a, and the cross product of a and v is b. Therefore, this is similar to saying that v is such that when you cross a with v, you get b. Since both a and b are fixed, and v is in the plane perpendicular to a.

Alternatively, since a × v = b, and v is perpendicular to a, then we can think of this as a linear equation in the plane. Let me think of the plane perpendicular to a. Let me denote that plane as P. Then, in this plane, the cross product a × v is equivalent to a scalar multiple of the normal vector of the plane, which is a. Wait, no. Wait, in the plane P, which is perpendicular to a, the cross product of a with any vector in P would be a vector in P? Wait, no. Wait, if v is in P, then a × v is perpendicular to both a and v. Since a is perpendicular to P, then a × v is in P? Wait, no. Wait, if a is perpendicular to P, then for any vector v in P, a × v is a vector that is perpendicular to both a and v. Since v is in P, which is perpendicular to a, then a × v is in the direction of the normal vector of P? Wait, no. Wait, if a is perpendicular to P, then the normal vector of P is a. Therefore, if v is in P, then a × v is a vector that is perpendicular to both a and v. Since a is perpendicular to v, then a × v is a vector that is in the plane P? Wait, no. Wait, if a is perpendicular to v, then a × v is a vector that is perpendicular to both a and v. Since a is perpendicular to v, then a × v is a vector that is in the plane spanned by a and v? Wait, no. Wait, the cross product of two vectors is perpendicular to both. So, if a is perpendicular to v, then a × v is perpendicular to both a and v. Therefore, since a is perpendicular to v, then a × v is in the plane spanned by a and v? No, because it's perpendicular to both. Wait, if you have two vectors a and v, then a × v is perpendicular to both. So, if a is perpendicular to v, then a × v is perpendicular to both, so it's in the direction of the normal vector of the plane spanned by a and v. But since v is in the plane P (perpendicular to a), then the plane spanned by a and v is the plane containing a and v, which is not necessarily the same as P. Wait, this is getting confusing. Maybe I should think in terms of coordinates.

Alternatively, since we already have a solution, and we confirmed that it's perpendicular to a, and that it gives the minimal magnitude, maybe this is sufficient. However, just to explore this idea, if we want to find v such that a × v = b and v is perpendicular to a, then we can use the fact that in this case, the cross product a × v = b implies that b is in the direction of the cross product, which is perpendicular to both a and v. But since v is perpendicular to a, then b is in the direction of the cross product of a and v, which is perpendicular to both. Therefore, since b is fixed, we can think of this as a system where we need to find v such that a × v = b and a · v = 0.

But we already have that from the cross product equation. Wait, no. Wait, if we have a × v = b, then since b is perpendicular to a, and v is perpendicular to a, then we can use the relation that a × v = b. But how does that help us?

Alternatively, since we have two equations:

1) a × v = b

2) a · v = 0

But in our case, the second equation is not necessarily given. However, in our solution, we found that v is perpendicular to a. Therefore, maybe we can use these two equations to solve for v. Let me try.

Let me denote a = (1, 2, -5), b = (90, 30, 30). Let me suppose that v is perpendicular to a, so a · v = 0. Then, we have a × v = b. Let me write these equations.

Let me denote v = (x, y, z). Then:

Equation 1: a · v = 0 ⇒ 1*x + 2*y -5*z = 0

Equation 2: a × v = b ⇒ (2z + 5y, -z -5x, y - 2x) = (90, 30, 30)

So, this is the same system as before, but with an additional equation. Wait, but in our previous approach, we had three equations, but they were dependent. However, here, we have two equations from the cross product and one from the dot product. Wait, but in reality, if we have a × v = b and a · v = 0, then we have three equations. Wait, but in our previous case, we had three equations, but they were dependent. However, here, if we have a × v = b and a · v = 0, then we have three equations. Let me check.

Wait, the cross product gives three equations, and the dot product gives one equation. However, in our previous case, the cross product equations were dependent. However, if we add the dot product equation, then we have four equations? Wait, no. Wait, the cross product gives three equations, and the dot product gives one equation. However, in reality, the cross product equations already include the condition that b is perpendicular to a, so maybe there is some dependency. Let me check.

But in our case, we already have that b is perpendicular to a, so the cross product equation is consistent. However, if we add the dot product equation, we have more constraints. Wait, but in our previous solution, we found that v is perpendicular to a. Therefore, maybe solving the cross product equations already gives us that v is perpendicular to a? Wait, no. Wait, in our previous solution, we had three equations from the cross product, which were not independent, and we found that v is perpendicular to a. Wait, but how?

Wait, let me check. From the cross product equations, we had:

Equation 1: 5y + 2z = 90

Equation 2: -5x - z = 30

Equation 3: -2x + y = 30

From equation 3: y = 2x + 30

From equation 2: z = -5x -30

Then, substituting into equation 1: 5*(2x + 30) + 2*(-5x -30) = 10x + 150 -10x -60 = 90. Which is 90 = 90. So, equation 1 is redundant. Therefore, the system has infinitely many solutions, parameterized by x, and in this case, the vector v is not necessarily perpendicular to a. Wait, but in our solution, we found that v is perpendicular to a. How come?

Wait, because when we found the vector v that minimizes the magnitude, we found that it is perpendicular to a. However, in general, the solutions to a × v = b are not necessarily perpendicular to a. Wait, but in our case, the minimal magnitude solution is perpendicular to a. So, perhaps the minimal magnitude solution is the one where v is perpendicular to a. Therefore, if we impose that v is perpendicular to a, then we can find a unique solution. Let me check.

So, if we have both a × v = b and a · v = 0, then we can solve for v. Let me try solving these equations.

So, equations:

1) 1*x + 2*y -5*z = 0 (from a · v = 0)

2) 2z + 5y = 90 (from cross product first component)

3) -z -5x = 30 (from cross product second component)

4) y - 2x = 30 (from cross product third component)

Wait, but equations 2, 3, 4 are the same as before. So, we have equations 2, 3, 4 and equation 1. However, in our previous solution, we had equations 2, 3, 4, which were dependent, and we found that v is parameterized by x. However, if we add equation 1, which is a · v = 0, then we can check if this is automatically satisfied.

Let me check with our solution v = (-7, 16, 5). Compute a · v = 1*(-7) + 2*16 + (-5)*5 = -7 + 32 -25 = 0. Yes, it is satisfied. Therefore, in this case, the solution that minimizes the magnitude automatically satisfies a · v = 0. Therefore, perhaps in general, the minimal magnitude solution is the one where v is perpendicular to a. Therefore, if we solve the system with a · v = 0 and a × v = b, we get a unique solution, which is the minimal one.

But how does that work? Because in our previous case, the system of equations from the cross product had infinitely many solutions, but when we impose the additional condition that v is perpendicular to a, we get a unique solution. However, in reality, the condition that v is perpendicular to a is not an additional equation, but rather a consequence of the minimal magnitude condition. Therefore, perhaps there's a relationship here.

Alternatively, since the minimal magnitude solution lies in the plane perpendicular to a, and since the cross product a × v = b is also in that plane, then there is a unique solution in that plane. Therefore, solving for v in that plane gives the unique solution. Therefore, this is another way to think about it.

But regardless of the approach, we have already found the solution, and verified it multiple times. Therefore, I think it's safe to conclude that the vector v with the smallest magnitude is (-7, 16, 5).

Therefore, the answer is \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.

**Final Answer**
The vector $\mathbf{v}$ with the smallest magnitude is \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.

To solve the problem, we are given the vector equation:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \mathbf{v} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}
$$

We are to find the vector $\mathbf{v}$ that satisfies this equation and has the **smallest magnitude**.

---

### Step 1: Express $\mathbf{v}$ in terms of components

Let $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$. Then, the cross product is:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix}
2z + 5y \\
-z - 5x \\
y - 2x
\end{pmatrix}
$$

Set this equal to the given vector:

$$
\begin{cases}
2z + 5y = 90 \\
-z - 5x = 30 \\
y - 2x = 30
\end{cases}
$$

---

### Step 2: Solve the system of equations

From the third equation:

$$
y = 2x + 30
$$

Substitute into the first equation:

$$
2z + 5(2x + 30) = 90 \Rightarrow 2z + 10x + 150 = 90 \Rightarrow 2z = -10x - 60 \Rightarrow z = -5x - 30
$$

Now we have:

$$
\mathbf{v} = \begin{pmatrix} x \\ 2x + 30 \\ -5x - 30 \end{pmatrix}
$$

---

### Step 3: Minimize the magnitude of $\mathbf{v}$

The magnitude squared of $\mathbf{v}$ is:

$$
||\mathbf{v}||^2 = x^2 + (2x + 30)^2 + (-5x - 30)^2
$$

Expand each term:

- $x^2$
- $(2x + 30)^2 = 4x^2 + 120x + 900$
- $(-5x - 30)^2 = 25x^2 + 300x + 900$

Add them:

$$
||\mathbf{v}||^2 = x^2 + 4x^2 + 120x + 900 + 25x^2 + 300x + 900 = 30x^2 + 420x + 1800
$$

This is a quadratic in $x$, and since the coefficient of $x^2$ is positive, the minimum occurs at the vertex:

$$
x = -\frac{b}{2a} = -\frac{420}{2 \cdot 30} = -7
$$

---

### Step 4: Find the corresponding $\mathbf{v}$

Substitute $x = -7$ into the expressions for $y$ and $z$:

- $y = 2(-7) + 30 = -14 + 30 = 16$
- $z = -5(-7) - 30 = 35 - 30 = 5$

So, the vector is:

$$
\mathbf{v} = \begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}
$$

---

### Step 5: Verify the solution

Compute the cross product:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix} =
\begin{pmatrix}
2(5) + 5(16) = 10 + 80 = 90 \\
-(1(5) + 5(-7)) = -(5 - 35) = 30 \\
1(16) - 2(-7) = 16 + 14 = 30
\end{pmatrix}
= \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}
$$

✅ The cross product matches the given vector.

---

### Step 6: Confirm minimal magnitude

The magnitude squared is:

$$
(-7)^2 + 16^2 + 5^2 = 49 + 256 + 25 = 330
$$

So the magnitude is $\sqrt{330}$, which is the **smallest possible** since the quadratic function is minimized at $x = -7$.

---

### Final Answer

$$
\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}
$$

[Assistant]
Explanation:
Okay, so I need to find a vector v such that when I take the cross product of the vector (1, 2, -5) with v, I get the vector (90, 30, 30). And among all possible such vectors v, I need the one with the smallest magnitude. Hmm, cross products... Let me recall how cross products work.

First, the cross product of two vectors in three dimensions results in a vector that's perpendicular to both of the original vectors. So, if I have vectors a and b, then a × b is perpendicular to both a and b. That means that the cross product here, which is (1, 2, -5) × v = (90, 30, 30), must be perpendicular to both (1, 2, -5) and v. Wait, but maybe that's not directly helpful here. Let me think.

Alternatively, maybe I can set up equations. Let me denote vector v as (x, y, z). Then, the cross product of (1, 2, -5) and (x, y, z) is given by the determinant of a matrix with the standard unit vectors i, j, k in the first row, the components of the first vector in the second row, and the components of the second vector in the third row. Let me write that out.

So, cross product:

|i   j   k |
|1   2  -5|
|x   y   z|

Which is i*(2*z - (-5)*y) - j*(1*z - (-5)*x) + k*(1*y - 2*x)

Simplifying each component:

First component (i): 2z + 5y

Second component (j): -(z + 5x) → -z -5x

Third component (k): y - 2x

So, the cross product is (2z + 5y, -z -5x, y - 2x). And this is equal to (90, 30, 30). Therefore, we can set up the following system of equations:

1) 2z + 5y = 90

2) -z -5x = 30

3) y - 2x = 30

So, three equations with three variables x, y, z. Let me write them again:

Equation 1: 5y + 2z = 90

Equation 2: -5x - z = 30

Equation 3: -2x + y = 30

So, I need to solve this system for x, y, z. Let me try solving equations 2 and 3 first, since equation 3 has only x and y, and equation 2 has x and z. Maybe I can express variables in terms of others.

From equation 3: y = 2x + 30. Let me write that down.

Equation 3: y = 2x + 30.

Then, substitute this into equation 1. Equation 1 is 5y + 2z = 90. So replacing y:

5*(2x + 30) + 2z = 90

Compute that:

10x + 150 + 2z = 90

Subtract 150 from both sides:

10x + 2z = 90 - 150 = -60

Divide both sides by 2:

5x + z = -30

So, equation 1 becomes 5x + z = -30. But equation 2 is -5x - z = 30. Wait, equation 2 is -5x - z = 30. Let me check:

Original equation 2: -z -5x = 30, which is same as -5x - z = 30. So, if I have equation 1 transformed as 5x + z = -30, which is the same as -5x - z = 30. Wait, that's exactly equation 2. So equations 1 and 2 are dependent? So, that means that equations 1 and 2 are not independent. Therefore, we have only two equations here, but three variables. Therefore, there are infinitely many solutions, which matches the problem statement that there are infinitely many vectors v. So, we need to find the vector v with the smallest magnitude.

So, since we have two equations and three variables, we can express two variables in terms of the third. Let me see. From equation 3, we have y = 2x + 30. From equation 2, which is equivalent to equation 1 transformed, we have 5x + z = -30. Therefore, z = -30 -5x. Therefore, we can express y and z in terms of x. Therefore, vector v is (x, y, z) = (x, 2x + 30, -5x -30). Therefore, v can be written as:

v = (x, 2x + 30, -5x -30)

So, now, we need to find the value of x that minimizes the magnitude of v. The magnitude squared of v is x² + (2x + 30)² + (-5x -30)². Since the square root is a monotonic function, minimizing the magnitude is equivalent to minimizing the magnitude squared. So, let me compute that.

Let me compute the magnitude squared:

||v||² = x² + (2x + 30)² + (-5x -30)²

Let me expand each term:

First term: x²

Second term: (2x + 30)² = 4x² + 120x + 900

Third term: (-5x -30)² = (5x + 30)² = 25x² + 300x + 900

So, adding all three terms:

x² + 4x² + 120x + 900 + 25x² + 300x + 900

Combine like terms:

x² + 4x² + 25x² = 30x²

120x + 300x = 420x

900 + 900 = 1800

Therefore, ||v||² = 30x² + 420x + 1800

So, we need to find the value of x that minimizes this quadratic function. Since it's a quadratic in x with a positive coefficient on x², the minimum occurs at the vertex. The x-coordinate of the vertex is at -b/(2a). Here, a = 30, b = 420. Therefore, x = -420/(2*30) = -420/60 = -7.

So, x = -7. Therefore, substituting back into the expressions for y and z:

y = 2x + 30 = 2*(-7) + 30 = -14 + 30 = 16

z = -5x -30 = -5*(-7) -30 = 35 - 30 = 5

Therefore, the vector v is (-7, 16, 5). Let me check if this is correct.

First, let me verify that this vector satisfies the original cross product equation.

Compute (1, 2, -5) × (-7, 16, 5):

Using the cross product formula:

i*(2*5 - (-5)*16) - j*(1*5 - (-5)*(-7)) + k*(1*16 - 2*(-7))

Compute each component:

First component (i): 2*5 - (-5)*16 = 10 + 80 = 90

Second component (j): -(1*5 - (-5)*(-7)) = -(5 - 35) = -(-30) = 30

Third component (k): 1*16 - 2*(-7) = 16 + 14 = 30

So, the cross product is (90, 30, 30), which matches the given vector. Great, so that checks out.

Now, let me check the magnitude of this vector. The magnitude squared is 30*(-7)^2 + 420*(-7) + 1800. Wait, but maybe I can compute it directly:

v = (-7, 16, 5). So, magnitude squared is (-7)^2 + 16^2 + 5^2 = 49 + 256 + 25 = 49 + 256 is 305, plus 25 is 330. So, magnitude is sqrt(330). Let me check if that's indeed the minimum.

Alternatively, since we found the minimum of the quadratic function, which was 30x² + 420x + 1800. Let me compute the value at x = -7:

30*(-7)^2 + 420*(-7) + 1800 = 30*49 - 2940 + 1800 = 1470 - 2940 + 1800 = (1470 + 1800) - 2940 = 3270 - 2940 = 330. Which matches. So, the minimum magnitude squared is 330, so the magnitude is sqrt(330). Therefore, this is the vector with the smallest magnitude.

But let me just think again if there's another way to approach this problem, maybe using vector projections or something else, to confirm.

Another approach: The cross product a × v = b. Then, we can think of this as a vector equation. The cross product a × v is a vector perpendicular to both a and v. However, since we have a specific b, we can think of this as solving for v. But since cross product is not invertible in general, we have infinitely many solutions, as we saw.

But perhaps we can use the formula for solving cross product equations. Let me recall that if a × v = b, then v can be expressed as v = (a × b)/||a||² * a + something? Wait, maybe not. Let me think.

Alternatively, since a × v = b, then v must be in the plane perpendicular to b, because the cross product is perpendicular to both a and v. Wait, no. Wait, the cross product a × v is perpendicular to both a and v. Therefore, the vector b must be perpendicular to a. Wait, is that true?

Wait, if a × v = b, then b is perpendicular to a, because cross product is perpendicular to both vectors. Therefore, if a × v = b, then b is perpendicular to a. Let me check if that's the case here. The given b is (90, 30, 30). Let me check if it's perpendicular to a = (1, 2, -5). The dot product is 1*90 + 2*30 + (-5)*30 = 90 + 60 - 150 = 0. Yes! So, the dot product is zero, so they are perpendicular. Therefore, that's consistent. So, that's a good check.

Therefore, since b is perpendicular to a, then the equation a × v = b has solutions. Now, how to find v?

Another approach: Let me recall that if a × v = b, then v can be written as v = (a × b)/||a||² * a + c, where c is any vector perpendicular to a. Wait, maybe? Let me think.

Alternatively, since a × v = b, then v can be expressed as v = (a × b)/||a||² * a + k, where k is a vector such that a · k = 0. Wait, maybe not. Let me think again.

Let me consider that if I have a × v = b, then v can be written as v = (a × b)/||a||² * a + t, where t is a vector such that a · t = 0. Wait, let me check.

Let me suppose that v = (a × b)/||a||² * a + t, where t is perpendicular to a. Then, compute a × v:

a × v = a × [ (a × b)/||a||² * a + t ] = (a × (a × b))/||a||² * a + a × t

But since t is perpendicular to a, then a × t is a vector perpendicular to a. However, the first term is (a × (a × b))/||a||² * a. Let me compute a × (a × b). Using the vector triple product formula: a × (a × b) = a(a · b) - b(a · a). But since a · b = 0, as we saw earlier, this simplifies to -b||a||². Therefore, substituting back:

a × v = [ -b||a||² / ||a||² ] * a + a × t = -b a + a × t

But we want a × v = b. Therefore, -b a + a × t = b. Therefore, a × t = b + b a. Wait, this seems complicated. Maybe this approach is not the best.

Alternatively, since we have a × v = b, and we know that v can be expressed as v = (a × b)/||a||² * a + c, where c is any vector perpendicular to a. Wait, let me check.

Let me suppose that v = (a × b)/||a||² * a + c, where c is perpendicular to a. Then, compute a × v:

a × v = a × [ (a × b)/||a||² * a + c ] = (a × (a × b))/||a||² * a + a × c

Again, using the triple product formula, a × (a × b) = a(a · b) - b(a · a). Since a · b = 0, this becomes -b||a||². Therefore:

a × v = [ -b||a||² / ||a||² ] * a + a × c = -b a + a × c

But we want this to be equal to b. Therefore:

-b a + a × c = b ⇒ a × c = b + b a

Wait, this seems messy. Maybe this approach isn't helpful. Let me try plugging in numbers.

Alternatively, since we have already solved the system of equations and found the vector v = (-7, 16, 5), which gives the minimal magnitude, maybe that's sufficient. However, let me think if there's another way to find the minimal magnitude.

Alternatively, since we have v expressed in terms of x as (x, 2x + 30, -5x -30), then the magnitude squared is 30x² + 420x + 1800. To find the minimum, we can take derivative with respect to x and set to zero. But since it's a quadratic, we already know the vertex is at x = -7. So, that's consistent.

Alternatively, maybe we can think geometrically. The set of all vectors v satisfying a × v = b is a line in 3D space. The minimal magnitude vector on that line is the one closest to the origin. Therefore, we can find the vector v on the line that is closest to the origin, which is the projection of the origin onto the line. But how to compute that?

Alternatively, since we have parametric equations for the line. Wait, the solutions for v are given by v = (x, 2x + 30, -5x -30). So, this is a line in parametric form with parameter x. So, we can write this as:

v = x*(1, 2, -5) + (0, 30, -30). Wait, let me check:

If x is multiplied by (1, 2, -5), then adding (0, 30, -30) gives (x, 2x + 30, -5x -30). Yes, that's correct. Therefore, the line is parametrized as v = x*(1, 2, -5) + (0, 30, -30). Therefore, the direction vector of the line is (1, 2, -5), and a point on the line is (0, 30, -30). Therefore, to find the vector on this line closest to the origin, we can use the projection formula.

The closest point on the line to the origin is the projection of the origin onto the line. Wait, but the line is given by v = t*(1, 2, -5) + (0, 30, -30), where t is a scalar parameter. Wait, but in our case, the parameter was x, but here I can use t as the parameter. So, the line is parametrized as:

v(t) = t*(1, 2, -5) + (0, 30, -30)

So, to find the value of t that minimizes ||v(t)||². Which is exactly what we did earlier, by substituting into the magnitude squared and finding the minimum. However, maybe we can use the projection formula here.

The formula for the projection of a vector onto a line. Let me recall that if we have a line L through point p with direction vector d, then the closest point on L to a point q is p + proj_d(q - p). Wait, but here, the line is through the point (0, 30, -30) with direction vector (1, 2, -5). The origin is the point q = (0, 0, 0). Therefore, the vector from p to q is q - p = (0, 0, 0) - (0, 30, -30) = (0, -30, 30). Then, the projection of this vector onto the direction vector d = (1, 2, -5) is:

proj_d(q - p) = [(q - p) · d / ||d||²] * d

Therefore, the closest point on the line to the origin is p + proj_d(q - p). Let me compute this.

First, compute (q - p) · d:

(0, -30, 30) · (1, 2, -5) = 0*1 + (-30)*2 + 30*(-5) = 0 - 60 - 150 = -210

Then, ||d||² = 1² + 2² + (-5)² = 1 + 4 + 25 = 30

Therefore, proj_d(q - p) = (-210 / 30) * d = -7 * (1, 2, -5) = (-7, -14, 35)

Therefore, the closest point on the line to the origin is p + proj_d(q - p) = (0, 30, -30) + (-7, -14, 35) = (0 -7, 30 -14, -30 +35) = (-7, 16, 5). Which is exactly the vector we found earlier! So, that confirms that the minimal magnitude vector is (-7, 16, 5). Therefore, this method also gives the same result. So, that's reassuring.

Therefore, regardless of the method, we get the same vector. Therefore, I can be confident that this is the correct answer.

But just to recap, the steps were:

1. Set up the cross product equation, leading to three equations.

2. Realize that there are infinitely many solutions, parameterized by x.

3. Express the vector v in terms of x.

4. Compute the magnitude squared as a function of x, find its minimum by completing the square or using vertex formula.

5. Alternatively, use projection formula to find the closest point on the line to the origin.

Either way, we arrive at the vector (-7, 16, 5). Let me just check once more the magnitude squared:

(-7)^2 + 16^2 + 5^2 = 49 + 256 + 25 = 49 + 256 is 305, plus 25 is 330. So sqrt(330) is approximately 18.165... Is there a possibility that there's a smaller magnitude? Well, since we found the minimum of the quadratic function, and the quadratic is convex, there can't be a smaller value. Therefore, this must be the minimal.

Alternatively, if I think about the cross product equation, the magnitude of the cross product is equal to ||a|| ||v|| sin(theta), where theta is the angle between a and v. Therefore, ||a × v|| = ||a|| ||v|| sin(theta). Therefore, given that ||a × v|| is fixed (which is ||(90, 30, 30)||), then ||v|| is minimized when sin(theta) is maximized, i.e., when theta = 90 degrees, so that sin(theta) = 1. Therefore, the minimal ||v|| occurs when v is perpendicular to a. Wait, but is that the case?

Wait, if theta is the angle between a and v, then ||a × v|| = ||a|| ||v|| sin(theta). Therefore, if we fix ||a × v||, then ||v|| is minimized when sin(theta) is maximized. Since sin(theta) ≤ 1, the maximum value is 1, which occurs when theta = 90 degrees. Therefore, the minimal ||v|| occurs when v is perpendicular to a. Therefore, if we can find such a vector v that is perpendicular to a and satisfies a × v = b, then that would be the minimal one.

But wait, in our case, we found that v = (-7, 16, 5). Let me check if this vector is perpendicular to a = (1, 2, -5). Compute their dot product:

(-7)(1) + 16*2 + 5*(-5) = -7 + 32 -25 = 0. Yes! So, the dot product is zero. Therefore, v is indeed perpendicular to a. Therefore, this is consistent with the above reasoning. Therefore, this is the minimal ||v||.

Therefore, this gives another way to think about the problem: find a vector v perpendicular to a such that a × v = b. Since when v is perpendicular to a, sin(theta) = 1, so ||v|| is minimized. Therefore, solving for such a v would give the minimal magnitude.

But how do we find such a v? Well, since v is perpendicular to a, then v lies in the plane perpendicular to a. Also, since a × v = b, which is perpendicular to a, as we saw earlier. Therefore, the vector b lies in the plane perpendicular to a, and v is also in that plane. Therefore, we can think of this as a problem in 2D: find v in the plane perpendicular to a such that a × v = b. But since we are in 3D, but the cross product in that plane would be equivalent to a 2D cross product.

Alternatively, since v is perpendicular to a, we can write v as a vector in the plane perpendicular to a. Let me try to express this.

Let me denote that since v is perpendicular to a, then v can be written as v = c × a, where c is some vector? Wait, no. Wait, if v is perpendicular to a, then v is in the plane perpendicular to a. However, the cross product a × v is a vector perpendicular to both a and v, which is in the direction of b. Wait, but since b is also perpendicular to a, then b is in the plane perpendicular to a. Therefore, the cross product a × v = b implies that v is in the plane perpendicular to a, and the cross product of a and v is b. Therefore, this is similar to saying that v is such that when you cross a with v, you get b. Since both a and b are fixed, and v is in the plane perpendicular to a.

Alternatively, since a × v = b, and v is perpendicular to a, then we can think of this as a linear equation in the plane. Let me think of the plane perpendicular to a. Let me denote that plane as P. Then, in this plane, the cross product a × v is equivalent to a scalar multiple of the normal vector of the plane, which is a. Wait, no. Wait, in the plane P, which is perpendicular to a, the cross product of a with any vector in P would be a vector in P? Wait, no. Wait, if v is in P, then a × v is perpendicular to both a and v. Since a is perpendicular to P, then a × v is in P? Wait, no. Wait, if a is perpendicular to P, then for any vector v in P, a × v is a vector that is perpendicular to both a and v. Since v is in P, which is perpendicular to a, then a × v is in the direction of the normal vector of P? Wait, no. Wait, if a is perpendicular to P, then the normal vector of P is a. Therefore, if v is in P, then a × v is a vector that is perpendicular to both a and v. Since a is perpendicular to v, then a × v is a vector that is in the plane P? Wait, no. Wait, if a is perpendicular to v, then a × v is a vector that is perpendicular to both a and v. Since a is perpendicular to v, then a × v is a vector that is in the plane spanned by a and v? Wait, no. Wait, the cross product of two vectors is perpendicular to both. So, if a is perpendicular to v, then a × v is perpendicular to both a and v. Therefore, since a is perpendicular to v, then a × v is in the plane spanned by a and v? No, because it's perpendicular to both. Wait, if you have two vectors a and v, then a × v is perpendicular to both. So, if a is perpendicular to v, then a × v is perpendicular to both, so it's in the direction of the normal vector of the plane spanned by a and v. But since v is in the plane P (perpendicular to a), then the plane spanned by a and v is the plane containing a and v, which is not necessarily the same as P. Wait, this is getting confusing. Maybe I should think in terms of coordinates.

Alternatively, since we already have a solution, and we confirmed that it's perpendicular to a, and that it gives the minimal magnitude, maybe this is sufficient. However, just to explore this idea, if we want to find v such that a × v = b and v is perpendicular to a, then we can use the fact that in this case, the cross product a × v = b implies that b is in the direction of the cross product, which is perpendicular to both a and v. But since v is perpendicular to a, then b is in the direction of the cross product of a and v, which is perpendicular to both. Therefore, since b is fixed, we can think of this as a system where we need to find v such that a × v = b and a · v = 0.

But we already have that from the cross product equation. Wait, no. Wait, if we have a × v = b, then since b is perpendicular to a, and v is perpendicular to a, then we can use the relation that a × v = b. But how does that help us?

Alternatively, since we have two equations:

1) a × v = b

2) a · v = 0

But in our case, the second equation is not necessarily given. However, in our solution, we found that v is perpendicular to a. Therefore, maybe we can use these two equations to solve for v. Let me try.

Let me denote a = (1, 2, -5), b = (90, 30, 30). Let me suppose that v is perpendicular to a, so a · v = 0. Then, we have a × v = b. Let me write these equations.

Let me denote v = (x, y, z). Then:

Equation 1: a · v = 0 ⇒ 1*x + 2*y -5*z = 0

Equation 2: a × v = b ⇒ (2z + 5y, -z -5x, y - 2x) = (90, 30, 30)

So, this is the same system as before, but with an additional equation. Wait, but in our previous approach, we had three equations, but they were dependent. However, here, we have two equations from the cross product and one from the dot product. Wait, but in reality, if we have a × v = b and a · v = 0, then we have three equations. Wait, but in our previous case, we had three equations, but they were dependent. However, here, if we have a × v = b and a · v = 0, then we have three equations. Let me check.

Wait, the cross product gives three equations, and the dot product gives one equation. However, in our previous case, the cross product equations were dependent. However, if we add the dot product equation, then we have four equations? Wait, no. Wait, the cross product gives three equations, and the dot product gives one equation. However, in reality, the cross product equations already include the condition that b is perpendicular to a, so maybe there is some dependency. Let me check.

But in our case, we already have that b is perpendicular to a, so the cross product equation is consistent. However, if we add the dot product equation, we have more constraints. Wait, but in our previous solution, we found that v is perpendicular to a. Therefore, maybe solving the cross product equations already gives us that v is perpendicular to a? Wait, no. Wait, in our previous solution, we had three equations from the cross product, which were not independent, and we found that v is perpendicular to a. Wait, but how?

Wait, let me check. From the cross product equations, we had:

Equation 1: 5y + 2z = 90

Equation 2: -5x - z = 30

Equation 3: -2x + y = 30

From equation 3: y = 2x + 30

From equation 2: z = -5x -30

Then, substituting into equation 1: 5*(2x + 30) + 2*(-5x -30) = 10x + 150 -10x -60 = 90. Which is 90 = 90. So, equation 1 is redundant. Therefore, the system has infinitely many solutions, parameterized by x, and in this case, the vector v is not necessarily perpendicular to a. Wait, but in our solution, we found that v is perpendicular to a. How come?

Wait, because when we found the vector v that minimizes the magnitude, we found that it is perpendicular to a. However, in general, the solutions to a × v = b are not necessarily perpendicular to a. Wait, but in our case, the minimal magnitude solution is perpendicular to a. So, perhaps the minimal magnitude solution is the one where v is perpendicular to a. Therefore, if we impose that v is perpendicular to a, then we can find a unique solution. Let me check.

So, if we have both a × v = b and a · v = 0, then we can solve for v. Let me try solving these equations.

So, equations:

1) 1*x + 2*y -5*z = 0 (from a · v = 0)

2) 2z + 5y = 90 (from cross product first component)

3) -z -5x = 30 (from cross product second component)

4) y - 2x = 30 (from cross product third component)

Wait, but equations 2, 3, 4 are the same as before. So, we have equations 2, 3, 4 and equation 1. However, in our previous solution, we had equations 2, 3, 4, which were dependent, and we found that v is parameterized by x. However, if we add equation 1, which is a · v = 0, then we can check if this is automatically satisfied.

Let me check with our solution v = (-7, 16, 5). Compute a · v = 1*(-7) + 2*16 + (-5)*5 = -7 + 32 -25 = 0. Yes, it is satisfied. Therefore, in this case, the solution that minimizes the magnitude automatically satisfies a · v = 0. Therefore, perhaps in general, the minimal magnitude solution is the one where v is perpendicular to a. Therefore, if we solve the system with a · v = 0 and a × v = b, we get a unique solution, which is the minimal one.

But how does that work? Because in our previous case, the system of equations from the cross product had infinitely many solutions, but when we impose the additional condition that v is perpendicular to a, we get a unique solution. However, in reality, the condition that v is perpendicular to a is not an additional equation, but rather a consequence of the minimal magnitude condition. Therefore, perhaps there's a relationship here.

Alternatively, since the minimal magnitude solution lies in the plane perpendicular to a, and since the cross product a × v = b is also in that plane, then there is a unique solution in that plane. Therefore, solving for v in that plane gives the unique solution. Therefore, this is another way to think about it.

But regardless of the approach, we have already found the solution, and verified it multiple times. Therefore, I think it's safe to conclude that the vector v with the smallest magnitude is (-7, 16, 5).

Therefore, the answer is \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.

**Final Answer**
The vector $\mathbf{v}$ with the smallest magnitude is \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.

To solve the problem, we are given the vector equation:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \mathbf{v} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}
$$

We are to find the vector $\mathbf{v}$ that satisfies this equation and has the **smallest magnitude**.

---

### Step 1: Express $\mathbf{v}$ in terms of components

Let $\mathbf{v} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$. Then, the cross product is:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} x \\ y \\ z \end{pmatrix} =
\begin{pmatrix}
2z + 5y \\
-z - 5x \\
y - 2x
\end{pmatrix}
$$

Set this equal to the given vector:

$$
\begin{cases}
2z + 5y = 90 \\
-z - 5x = 30 \\
y - 2x = 30
\end{cases}
$$

---

### Step 2: Solve the system of equations

From the third equation:

$$
y = 2x + 30
$$

Substitute into the first equation:

$$
2z + 5(2x + 30) = 90 \Rightarrow 2z + 10x + 150 = 90 \Rightarrow 2z = -10x - 60 \Rightarrow z = -5x - 30
$$

Now we have:

$$
\mathbf{v} = \begin{pmatrix} x \\ 2x + 30 \\ -5x - 30 \end{pmatrix}
$$

---

### Step 3: Minimize the magnitude of $\mathbf{v}$

The magnitude squared of $\mathbf{v}$ is:

$$
||\mathbf{v}||^2 = x^2 + (2x + 30)^2 + (-5x - 30)^2
$$

Expand each term:

- $x^2$
- $(2x + 30)^2 = 4x^2 + 120x + 900$
- $(-5x - 30)^2 = 25x^2 + 300x + 900$

Add them:

$$
||\mathbf{v}||^2 = x^2 + 4x^2 + 120x + 900 + 25x^2 + 300x + 900 = 30x^2 + 420x + 1800
$$

This is a quadratic in $x$, and since the coefficient of $x^2$ is positive, the minimum occurs at the vertex:

$$
x = -\frac{b}{2a} = -\frac{420}{2 \cdot 30} = -7
$$

---

### Step 4: Find the corresponding $\mathbf{v}$

Substitute $x = -7$ into the expressions for $y$ and $z$:

- $y = 2(-7) + 30 = -14 + 30 = 16$
- $z = -5(-7) - 30 = 35 - 30 = 5$

So, the vector is:

$$
\mathbf{v} = \begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}
$$

---

### Step 5: Verify the solution

Compute the cross product:

$$
\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix} =
\begin{pmatrix}
2(5) + 5(16) = 10 + 80 = 90 \\
-(1(5) + 5(-7)) = -(5 - 35) = 30 \\
1(16) - 2(-7) = 16 + 14 = 30
\end{pmatrix}
= \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}
$$

✅ The cross product matches the given vector.

---

### Step 6: Confirm minimal magnitude

The magnitude squared is:

$$
(-7)^2 + 16^2 + 5^2 = 49 + 256 + 25 = 330
$$

So the magnitude is $\sqrt{330}$, which is the **smallest possible** since the quadratic function is minimized at $x = -7$.

---

### Final Answer

$$
\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}
$$

Solution:
Let $\mathbf{v} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}.$  Then
\[\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix},\]so
\[\begin{pmatrix} 5b + 2c \\ -5a - c \\ -2a + b \end{pmatrix} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}.\]Comparing the components, we get
\begin{align*}
5b + 2c &= 90, \\
-5a - c &= 30, \\
-2a + b &= 30.
\end{align*}From the second equation, $c = -5a - 30.$  From the third equation, $b = 2a + 30.$  We want to minimize the magnitude of $\mathbf{v},$ which is equivalent to minimizing
\[a^2 + b^2 + c^2 = a^2 + (2a + 30)^2 + (-5a - 30)^2 = 30a^2 + 420a + 1800.\]Completing the square, we get $30 (a + 7)^2 + 330,$ so the magnitude is minimized when $a = -7.$  Then $b = 16$ and $c = 5,$ so the vector $\mathbf{v}$ we seek is $\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.$


Explanation:
Okay, so I need to find the minimum value of the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ given that $a + b + c + d = 1$ where all variables are positive real numbers. Hmm, let me think about how to approach this.

First, I remember that optimization problems with constraints often use methods like Lagrange multipliers, but since this is an algebra problem, maybe there's a way to apply inequalities like Cauchy-Schwarz or AM-HM or AM-GM. Let me recall those.

The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can apply that here. But the problem is that the expression we need to minimize has different coefficients in the denominators. So maybe I need to adjust the variables or split them in a way that allows me to apply AM-GM effectively.

Alternatively, maybe I can use the method of Lagrange multipliers. Let me try that first. Let me set up the function to minimize:

$f(a, b, c, d) = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$

subject to the constraint $g(a, b, c, d) = a + b + c + d - 1 = 0$.

The method of Lagrange multipliers says that at the minimum, the gradient of f is proportional to the gradient of g. So:

$\nabla f = \lambda \nabla g$

Calculating the partial derivatives:

For a:
$\frac{\partial f}{\partial a} = -\frac{1}{a^2}$
Similarly for b: $-\frac{1}{b^2}$
For c: $-\frac{4}{c^2}$
For d: $-\frac{16}{d^2}$

Gradient of g is (1, 1, 1, 1). Therefore, setting up equations:

$-\frac{1}{a^2} = \lambda$
$-\frac{1}{b^2} = \lambda$
$-\frac{4}{c^2} = \lambda$
$-\frac{16}{d^2} = \lambda$

So from the first two equations, since both equal to λ, we have:

$-\frac{1}{a^2} = -\frac{1}{b^2}$ ⇒ $a^2 = b^2$ ⇒ since a and b are positive, a = b.

Similarly, from the third equation: $-\frac{4}{c^2} = \lambda$, and from the first equation, $-\frac{1}{a^2} = \lambda$. Therefore:

$-\frac{4}{c^2} = -\frac{1}{a^2}$ ⇒ $\frac{4}{c^2} = \frac{1}{a^2}$ ⇒ $4a^2 = c^2$ ⇒ $c = 2a$ (since c is positive).

Similarly, from the fourth equation: $-\frac{16}{d^2} = \lambda = -\frac{1}{a^2}$ ⇒ $\frac{16}{d^2} = \frac{1}{a^2}$ ⇒ $16a^2 = d^2$ ⇒ $d = 4a$.

So now, we have relations between variables:

a = b, c = 2a, d = 4a.

Now, substitute into the constraint equation:

a + b + c + d = 1 ⇒ a + a + 2a + 4a = 1 ⇒ 8a = 1 ⇒ a = 1/8.

Therefore:

a = b = 1/8,

c = 2*(1/8) = 1/4,

d = 4*(1/8) = 1/2.

Now, let's compute the value of the expression:

$\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$

Substituting the values:

$\frac{1}{1/8} + \frac{1}{1/8} + \frac{4}{1/4} + \frac{16}{1/2}$

Calculating each term:

1/(1/8) = 8,

Similarly, the second term is also 8,

Third term: 4 divided by 1/4 is 4 * 4 = 16,

Fourth term: 16 divided by 1/2 is 16 * 2 = 32.

Adding them up: 8 + 8 + 16 + 32 = 64.

So according to this method, the minimum value is 64. But wait, let me check if this is correct. Maybe I made a mistake in the calculation?

Wait, let me verify the substitution again:

a = 1/8, so 1/a = 8. Similarly for b. Then c = 1/4, so 4/c = 4/(1/4) = 16. d = 1/2, so 16/d = 16/(1/2) = 32. So total is 8 + 8 + 16 + 32 = 64. Yes, that seems correct.

But let me check if there's another way to approach this problem, maybe using Cauchy-Schwarz or AM-GM, to confirm.

Alternatively, maybe I can use the Cauchy-Schwarz inequality. Let me recall that for vectors u and v, (u ⋅ v)^2 ≤ (u ⋅ u)(v ⋅ v). But I'm not sure how to apply it here directly.

Alternatively, maybe I can use the method of weighted AM-HM inequality. Let me think.

Alternatively, since the problem has different coefficients in the denominators, maybe I can split the variables into groups. For example, since the denominators have coefficients 1, 1, 4, 16, which are squares of 1, 1, 2, 4. Wait, 1, 1, 2, 4... Maybe there's a pattern here. Let me think.

Alternatively, maybe I can use the Cauchy-Schwarz inequality in the following way. Let me consider the terms $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$. Let me think of this as $\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}$. Then, maybe apply Cauchy-Schwarz.

Recall that Cauchy-Schwarz inequality states that $(\sum \frac{x_i^2}{a_i}) \ge \frac{(\sum x_i)^2}{\sum a_i}$. Wait, if I set up the terms as $\frac{x_i^2}{a_i}$, then the sum would be greater or equal to $\frac{(\sum x_i)^2}{\sum a_i}$. Let me try that.

Let me set $x_1 = 1$, $x_2 = 1$, $x_3 = 2$, $x_4 = 4$, and $a_1 = a$, $a_2 = b$, $a_3 = c$, $a_4 = d$. Then, according to Cauchy-Schwarz:

$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \ge \frac{(1 + 1 + 2 + 4)^2}{a + b + c + d}
$$

Calculating numerator: (1 + 1 + 2 + 4) = 8, so squared is 64. Denominator is 1. Therefore, the inequality gives:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge \frac{64}{1} = 64
$$

So this gives the same result as the Lagrange multipliers method. Therefore, the minimum value is 64, achieved when equality holds in Cauchy-Schwarz. Equality holds when $\frac{x_1}{a_1} = \frac{x_2}{a_2} = \frac{x_3}{a_3} = \frac{x_4}{a_4}$, that is:

$\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}$

Which gives the same relations as before:

From $\frac{1}{a} = \frac{1}{b}$ ⇒ a = b.

From $\frac{1}{a} = \frac{2}{c}$ ⇒ c = 2a.

From $\frac{1}{a} = \frac{4}{d}$ ⇒ d = 4a.

Which is exactly the same as in the Lagrange multipliers method. Therefore, substituting back into the constraint gives a = 1/8, etc., leading to the same value of 64. Therefore, this seems to confirm that 64 is indeed the minimum.

But let me check if there's another approach, maybe using AM-GM. Let me try that.

The problem is to minimize $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ with a + b + c + d = 1.

If I consider AM-GM, I need to relate the terms. However, since the variables are in denominators, it's tricky. Maybe I can split the variables into parts. For example, since the coefficients in the denominators are 1, 1, 4, 16, which are 1^2, 1^2, 2^2, 4^2. Wait, maybe I can think of this as having different weights.

Alternatively, maybe I can use the method of substitution. Since a + b + c + d = 1, maybe express variables in terms of others. But that might complicate.

Alternatively, think of the expression as sum of terms and try to apply AM-GM on each term. But since they are reciprocals, it's not straightforward.

Alternatively, maybe split the variables into groups. For example, since we have two terms with 1/a and 1/b, maybe consider them together. Let me think.

Let me consider variables a and b. Let me assume that a = b, as found earlier. Then, maybe the minimum occurs when a = b. Similarly, c and d are related to a. But this is similar to the Lagrange multipliers approach.

Alternatively, maybe use AM-GM on the entire expression. Let me think. Suppose I consider the terms $\frac{1}{a}$, $\frac{1}{b}$, $\frac{4}{c}$, $\frac{16}{d}$. Let me think of them as four terms. But how to relate them to the sum a + b + c + d = 1.

Alternatively, use the Cauchy-Schwarz inequality as before, which already gave the answer. Since both methods gave me 64, and the equality condition is satisfied, I think that 64 is indeed the minimum.

But just to be thorough, let me test with some numbers. Suppose I take a = b = 1/8, c = 1/4, d = 1/2. Then the sum is 1/8 + 1/8 + 1/4 + 1/2 = 1/8 + 1/8 + 2/8 + 4/8 = 8/8 = 1. Correct. And the expression is 8 + 8 + 16 + 32 = 64. Now, let me try another set of numbers to see if I can get a lower value.

Suppose I take a = 0.2, b = 0.2, c = 0.2, d = 0.4. Then the sum is 0.2 + 0.2 + 0.2 + 0.4 = 1. Then compute the expression:

1/0.2 + 1/0.2 + 4/0.2 + 16/0.4 = 5 + 5 + 20 + 40 = 70, which is higher than 64. So worse.

Another test: suppose I take a = 0.1, b = 0.1, c = 0.2, d = 0.6. Then sum is 0.1 + 0.1 + 0.2 + 0.6 = 1. The expression is 1/0.1 + 1/0.1 + 4/0.2 + 16/0.6 = 10 + 10 + 20 + 26.666... ≈ 66.666..., which is still higher than 64.

Another test: suppose I take a = 1/8, b = 1/8, c = 1/4, d = 1/2. Which gives 64. What if I take a slightly different value? Let me try a = 0.12, b = 0.12, then c = 2a = 0.24, d = 4a = 0.48. Then sum is 0.12 + 0.12 + 0.24 + 0.48 = 0.96. Wait, that's less than 1. So need to adjust. Let me make a = 0.125, which is 1/8, then same as before. If I take a = 0.1, then c = 0.2, d = 0.4, and then b = ?

Wait, if I take a = 0.1, then since a = b, b = 0.1. Then c = 0.2, d = 0.4. Then total sum is 0.1 + 0.1 + 0.2 + 0.4 = 0.8. So need to add 0.2 more. Maybe adjust variables. Suppose I take a = 0.1 + x, b = 0.1 + x, c = 0.2 + y, d = 0.4 + z, such that (0.1 + x) + (0.1 + x) + (0.2 + y) + (0.4 + z) = 1. Which simplifies to 0.8 + 2x + y + z = 1 ⇒ 2x + y + z = 0.2. But this might complicate. Maybe not the best approach.

Alternatively, take a = 0.1, b = 0.1, c = 0.2, d = 0.6. Wait, that was tried before. The value was higher. So maybe 64 is indeed the minimum.

Alternatively, let me try to use AM-GM on the terms. Let me think. Suppose I consider the terms $\frac{1}{a}$, $\frac{1}{b}$, $\frac{4}{c}$, $\frac{16}{d}$. Let me think of them as four terms. Let me try to apply AM-GM on these four terms. The AM-GM inequality says that the arithmetic mean is greater or equal to the geometric mean. So:

$$
\frac{\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}}{4} \ge \sqrt[4]{\frac{1}{a} \cdot \frac{1}{b} \cdot \frac{4}{c} \cdot \frac{16}{d}} = \sqrt[4]{\frac{64}{abcd}}
$$

Therefore,

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge 4 \sqrt[4]{\frac{64}{abcd}} = 4 \cdot \left( \frac{64}{abcd} \right)^{1/4}
$$

But this seems not directly helpful because we don't know the value of abcd. However, maybe we can relate abcd to the constraint a + b + c + d = 1. But with AM-GM on a + b + c + d, we have:

$$
\frac{a + b + c + d}{4} \ge \sqrt[4]{abcd} \Rightarrow \frac{1}{4} \ge \sqrt[4]{abcd} \Rightarrow abcd \le \left( \frac{1}{4} \right)^4 = \frac{1}{256}
$$

Therefore, substituting back into the previous inequality:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge 4 \cdot \left( \frac{64}{abcd} \right)^{1/4} \ge 4 \cdot \left( \frac{64}{1/256} \right)^{1/4} = 4 \cdot \left( 64 \times 256 \right)^{1/4}
$$

Calculating 64 × 256: 64 × 256 = 16384. Then, 16384^(1/4). Let me compute that. 16384 is 2^14, since 2^10 = 1024, 2^14 = 16384. Therefore, 2^14^(1/4) = 2^(14/4) = 2^(3.5) = 2^3 * 2^0.5 = 8 * sqrt(2) ≈ 11.3137. Therefore, the lower bound would be 4 * 11.3137 ≈ 45.25. But this is way lower than 64, which suggests that this approach is not tight. Therefore, the AM-GM approach here is not giving a useful bound because the equality conditions are not met. Therefore, this method isn't helpful here. So, the previous methods of Cauchy-Schwarz and Lagrange multipliers are better.

Alternatively, maybe I can use the method of substitution. Let me try to express variables in terms of a single variable. Since from the Lagrange multipliers, we found that a = b, c = 2a, d = 4a. Therefore, the variables are dependent. Therefore, the minimal value occurs when variables are in this proportion. Therefore, the minimal value is 64.

Alternatively, maybe I can think of the problem as optimizing with weights. Let me think of the expression as $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$. Let me consider variables x = a, y = b, z = c, w = d. Then, the problem is to minimize $\frac{1}{x} + \frac{1}{y} + \frac{4}{z} + \frac{16}{w}$ with x + y + z + w = 1.

Alternatively, maybe I can use the method of substitution by letting variables be proportional to the coefficients in the denominators. Wait, in the Lagrange multipliers method, we found that the ratios between variables are determined by the coefficients in the expression. For example, since the partial derivatives are proportional to the coefficients in the denominators.

Alternatively, think of the problem as minimizing the sum of terms with different weights. Maybe using the concept of convexity. Since the function is convex, the minimum is achieved at the critical point found by Lagrange multipliers.

Alternatively, since both methods gave me 64, and testing with other values gives higher results, I think that 64 is indeed the correct minimum. Therefore, the answer is 64.

But just to make sure, let me try another set of variables. Suppose I take a = 1/8, b = 1/8, c = 1/4, d = 1/2. Then, if I slightly perturb one variable, say increase a by ε, then decrease another variable to keep the sum 1. Let me try:

Let me take a = 1/8 + ε, b = 1/8 - ε, c = 1/4, d = 1/2. Then, the sum is (1/8 + ε) + (1/8 - ε) + 1/4 + 1/2 = 1/8 + 1/8 + 1/4 + 1/2 = same as before, 1. Now compute the expression:

1/a + 1/b + 4/c + 16/d.

So 1/(1/8 + ε) + 1/(1/8 - ε) + 4/(1/4) + 16/(1/2).

Compute the first two terms:

Let me approximate for small ε. Let me take ε approaching 0.

1/(1/8 + ε) ≈ 8 - 64ε (using the expansion 1/(a + ε) ≈ 1/a - ε/a² for small ε)

Similarly, 1/(1/8 - ε) ≈ 8 + 64ε.

Adding them together: (8 - 64ε) + (8 + 64ε) = 16. So the first two terms sum to 16, same as before. The other terms are 16 + 32 = 48. So total is 16 + 48 = 64. Therefore, even after perturbation, the value remains the same? Wait, that's strange. Wait, but if I take ε ≠ 0, then the first two terms would be 1/(1/8 + ε) + 1/(1/8 - ε). Let me compute exactly:

Let me compute 1/(1/8 + ε) + 1/(1/8 - ε) = [ (1/8 - ε) + (1/8 + ε) ] / [ (1/8 + ε)(1/8 - ε) ] = (2/8) / [ (1/64 - ε²) ] = (1/4) / (1/64 - ε²) = (1/4) * [1 / (1/64 - ε²)].

Which is equal to (1/4) * [ 64 / (1 - 64 ε²) ] = 16 / (1 - 64 ε²) ≈ 16(1 + 64 ε²) for small ε. Therefore, the first two terms are approximately 16 + 16 * 64 ε². Therefore, the total expression is approximately 16 + 16 * 64 ε² + 48 = 64 + 1024 ε², which is greater than 64. Therefore, perturbing a and b slightly increases the value. Therefore, this suggests that the critical point is indeed a minimum.

Similarly, if I perturb other variables, say, increase c and decrease d, keeping the sum same. Let me try. Let me take c = 1/4 + ε, d = 1/2 - ε. Then, the sum remains 1. Compute the expression:

1/a + 1/b + 4/c + 16/d. Since a and b are still 1/8, so 8 + 8 = 16. Then 4/c = 4/(1/4 + ε) ≈ 16 - 64 ε (using similar expansion). And 16/d = 16/(1/2 - ε) ≈ 32 + 128 ε. Therefore, total expression is 16 + (16 - 64 ε) + (32 + 128 ε) = 16 + 16 + 32 + ( -64 ε + 128 ε ) = 64 + 64 ε. Therefore, if ε is positive, the value increases, and if ε is negative, the value decreases? Wait, but if ε is negative, then c = 1/4 + ε would be less than 1/4, and d = 1/2 - ε would be more than 1/2. Let me check with ε negative. Let me take ε = -k, where k > 0. Then:

4/c = 4/(1/4 - k) ≈ 16 + 64k (since 1/(1/4 - k) ≈ 4/(1 - 4k) ≈ 4(1 + 4k) = 4 + 16k, so 4/(1/4 - k) ≈ 16 + 64k? Wait, maybe my previous expansion was wrong. Let me recast:

Let me compute 4/c where c = 1/4 + ε. Let me write c = 1/4 + ε. Then 4/c = 4 / (1/4 + ε) = 16 / (1 + 4ε) ≈ 16(1 - 4ε + 16ε² - ...) ≈ 16 - 64ε + ... Similarly, 16/d where d = 1/2 - ε. Then 16/d = 32 / (1 - 2ε) ≈ 32(1 + 2ε + 4ε² + ...) ≈ 32 + 64ε + ... Therefore, adding them together: (16 - 64ε) + (32 + 64ε) = 48. Therefore, the total expression is 16 + 48 = 64. Wait, so if I take ε positive or negative, the first order terms cancel out? Wait, but earlier when I considered the expansion, I thought there was a term with ε. Maybe my previous calculation was wrong. Let me check again.

Wait, if I take c = 1/4 + ε, then 4/c = 4 / (1/4 + ε) = 16 / (1 + 4ε) ≈ 16(1 - 4ε + 16ε² - ...) ≈ 16 - 64ε + ... Similarly, d = 1/2 - ε, so 16/d = 16 / (1/2 - ε) = 32 / (1 - 2ε) ≈ 32(1 + 2ε + 4ε² + ...) ≈ 32 + 64ε + ... Therefore, adding these two terms: (16 - 64ε) + (32 + 64ε) = 48. Therefore, the first order terms cancel. Therefore, the total expression is 16 + 48 = 64, same as before. Therefore, the second order terms would be 16*(16ε²) + 32*(4ε²) = 256ε² + 128ε² = 384ε². Therefore, the total expression is 64 + 384ε², which is greater than 64. Therefore, perturbing c and d also leads to an increase in the value. Therefore, this suggests that the critical point is indeed a minimum.

Therefore, all these tests confirm that 64 is the minimum value. Therefore, I can be confident that the answer is 64.

**Final Answer**
\boxed{64}

To find the minimum value of the expression
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}
$$
under the constraint
$$
a + b + c + d = 1
$$
with $ a, b, c, d > 0 $, we can use the **Cauchy-Schwarz inequality** in a strategic way.

---

### Step 1: Apply Cauchy-Schwarz Inequality

We rewrite the expression as:
$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}
$$

Now apply the **Cauchy-Schwarz inequality** in the form:
$$
\left( \sum \frac{x_i^2}{a_i} \right) \ge \frac{(\sum x_i)^2}{\sum a_i}
$$

Let:
- $ x_1 = 1, x_2 = 1, x_3 = 2, x_4 = 4 $
- $ a_1 = a, a_2 = b, a_3 = c, a_4 = d $

Then:
$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \ge \frac{(1 + 1 + 2 + 4)^2}{a + b + c + d}
$$

Compute:
- Numerator: $ (1 + 1 + 2 + 4)^2 = 8^2 = 64 $
- Denominator: $ a + b + c + d = 1 $

So:
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge \frac{64}{1} = 64
$$

---

### Step 2: Equality Condition

Equality in Cauchy-Schwarz holds when:
$$
\frac{x_1}{a_1} = \frac{x_2}{a_2} = \frac{x_3}{a_3} = \frac{x_4}{a_4}
$$

That is:
$$
\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}
$$

From this, we deduce:
- $ a = b $
- $ c = 2a $
- $ d = 4a $

Substitute into the constraint:
$$
a + b + c + d = a + a + 2a + 4a = 8a = 1 \Rightarrow a = \frac{1}{8}
$$

Then:
- $ a = b = \frac{1}{8} $
- $ c = \frac{1}{4} $
- $ d = \frac{1}{2} $

---

### Step 3: Compute the Expression

Now compute:
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}
= \frac{1}{1/8} + \frac{1}{1/8} + \frac{4}{1/4} + \frac{16}{1/2}
= 8 + 8 + 16 + 32 = 64
$$

---

### Step 4: Confirm Minimum

We have shown that the expression is **greater than or equal to 64** by Cauchy-Schwarz, and that the value **64 is achievable** when the variables are set as above. Therefore, the **minimum value** of the expression is:

$$
\boxed{64}
$$

[Assistant]
Explanation:
Okay, so I need to find the minimum value of the expression $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ given that $a + b + c + d = 1$ where all variables are positive real numbers. Hmm, let me think about how to approach this.

First, I remember that optimization problems with constraints often use methods like Lagrange multipliers, but since this is an algebra problem, maybe there's a way to apply inequalities like Cauchy-Schwarz or AM-HM or AM-GM. Let me recall those.

The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Maybe I can apply that here. But the problem is that the expression we need to minimize has different coefficients in the denominators. So maybe I need to adjust the variables or split them in a way that allows me to apply AM-GM effectively.

Alternatively, maybe I can use the method of Lagrange multipliers. Let me try that first. Let me set up the function to minimize:

$f(a, b, c, d) = \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$

subject to the constraint $g(a, b, c, d) = a + b + c + d - 1 = 0$.

The method of Lagrange multipliers says that at the minimum, the gradient of f is proportional to the gradient of g. So:

$\nabla f = \lambda \nabla g$

Calculating the partial derivatives:

For a:
$\frac{\partial f}{\partial a} = -\frac{1}{a^2}$
Similarly for b: $-\frac{1}{b^2}$
For c: $-\frac{4}{c^2}$
For d: $-\frac{16}{d^2}$

Gradient of g is (1, 1, 1, 1). Therefore, setting up equations:

$-\frac{1}{a^2} = \lambda$
$-\frac{1}{b^2} = \lambda$
$-\frac{4}{c^2} = \lambda$
$-\frac{16}{d^2} = \lambda$

So from the first two equations, since both equal to λ, we have:

$-\frac{1}{a^2} = -\frac{1}{b^2}$ ⇒ $a^2 = b^2$ ⇒ since a and b are positive, a = b.

Similarly, from the third equation: $-\frac{4}{c^2} = \lambda$, and from the first equation, $-\frac{1}{a^2} = \lambda$. Therefore:

$-\frac{4}{c^2} = -\frac{1}{a^2}$ ⇒ $\frac{4}{c^2} = \frac{1}{a^2}$ ⇒ $4a^2 = c^2$ ⇒ $c = 2a$ (since c is positive).

Similarly, from the fourth equation: $-\frac{16}{d^2} = \lambda = -\frac{1}{a^2}$ ⇒ $\frac{16}{d^2} = \frac{1}{a^2}$ ⇒ $16a^2 = d^2$ ⇒ $d = 4a$.

So now, we have relations between variables:

a = b, c = 2a, d = 4a.

Now, substitute into the constraint equation:

a + b + c + d = 1 ⇒ a + a + 2a + 4a = 1 ⇒ 8a = 1 ⇒ a = 1/8.

Therefore:

a = b = 1/8,

c = 2*(1/8) = 1/4,

d = 4*(1/8) = 1/2.

Now, let's compute the value of the expression:

$\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$

Substituting the values:

$\frac{1}{1/8} + \frac{1}{1/8} + \frac{4}{1/4} + \frac{16}{1/2}$

Calculating each term:

1/(1/8) = 8,

Similarly, the second term is also 8,

Third term: 4 divided by 1/4 is 4 * 4 = 16,

Fourth term: 16 divided by 1/2 is 16 * 2 = 32.

Adding them up: 8 + 8 + 16 + 32 = 64.

So according to this method, the minimum value is 64. But wait, let me check if this is correct. Maybe I made a mistake in the calculation?

Wait, let me verify the substitution again:

a = 1/8, so 1/a = 8. Similarly for b. Then c = 1/4, so 4/c = 4/(1/4) = 16. d = 1/2, so 16/d = 16/(1/2) = 32. So total is 8 + 8 + 16 + 32 = 64. Yes, that seems correct.

But let me check if there's another way to approach this problem, maybe using Cauchy-Schwarz or AM-GM, to confirm.

Alternatively, maybe I can use the Cauchy-Schwarz inequality. Let me recall that for vectors u and v, (u ⋅ v)^2 ≤ (u ⋅ u)(v ⋅ v). But I'm not sure how to apply it here directly.

Alternatively, maybe I can use the method of weighted AM-HM inequality. Let me think.

Alternatively, since the problem has different coefficients in the denominators, maybe I can split the variables into groups. For example, since the denominators have coefficients 1, 1, 4, 16, which are squares of 1, 1, 2, 4. Wait, 1, 1, 2, 4... Maybe there's a pattern here. Let me think.

Alternatively, maybe I can use the Cauchy-Schwarz inequality in the following way. Let me consider the terms $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$. Let me think of this as $\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}$. Then, maybe apply Cauchy-Schwarz.

Recall that Cauchy-Schwarz inequality states that $(\sum \frac{x_i^2}{a_i}) \ge \frac{(\sum x_i)^2}{\sum a_i}$. Wait, if I set up the terms as $\frac{x_i^2}{a_i}$, then the sum would be greater or equal to $\frac{(\sum x_i)^2}{\sum a_i}$. Let me try that.

Let me set $x_1 = 1$, $x_2 = 1$, $x_3 = 2$, $x_4 = 4$, and $a_1 = a$, $a_2 = b$, $a_3 = c$, $a_4 = d$. Then, according to Cauchy-Schwarz:

$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \ge \frac{(1 + 1 + 2 + 4)^2}{a + b + c + d}
$$

Calculating numerator: (1 + 1 + 2 + 4) = 8, so squared is 64. Denominator is 1. Therefore, the inequality gives:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge \frac{64}{1} = 64
$$

So this gives the same result as the Lagrange multipliers method. Therefore, the minimum value is 64, achieved when equality holds in Cauchy-Schwarz. Equality holds when $\frac{x_1}{a_1} = \frac{x_2}{a_2} = \frac{x_3}{a_3} = \frac{x_4}{a_4}$, that is:

$\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}$

Which gives the same relations as before:

From $\frac{1}{a} = \frac{1}{b}$ ⇒ a = b.

From $\frac{1}{a} = \frac{2}{c}$ ⇒ c = 2a.

From $\frac{1}{a} = \frac{4}{d}$ ⇒ d = 4a.

Which is exactly the same as in the Lagrange multipliers method. Therefore, substituting back into the constraint gives a = 1/8, etc., leading to the same value of 64. Therefore, this seems to confirm that 64 is indeed the minimum.

But let me check if there's another approach, maybe using AM-GM. Let me try that.

The problem is to minimize $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$ with a + b + c + d = 1.

If I consider AM-GM, I need to relate the terms. However, since the variables are in denominators, it's tricky. Maybe I can split the variables into parts. For example, since the coefficients in the denominators are 1, 1, 4, 16, which are 1^2, 1^2, 2^2, 4^2. Wait, maybe I can think of this as having different weights.

Alternatively, maybe I can use the method of substitution. Since a + b + c + d = 1, maybe express variables in terms of others. But that might complicate.

Alternatively, think of the expression as sum of terms and try to apply AM-GM on each term. But since they are reciprocals, it's not straightforward.

Alternatively, maybe split the variables into groups. For example, since we have two terms with 1/a and 1/b, maybe consider them together. Let me think.

Let me consider variables a and b. Let me assume that a = b, as found earlier. Then, maybe the minimum occurs when a = b. Similarly, c and d are related to a. But this is similar to the Lagrange multipliers approach.

Alternatively, maybe use AM-GM on the entire expression. Let me think. Suppose I consider the terms $\frac{1}{a}$, $\frac{1}{b}$, $\frac{4}{c}$, $\frac{16}{d}$. Let me think of them as four terms. But how to relate them to the sum a + b + c + d = 1.

Alternatively, use the Cauchy-Schwarz inequality as before, which already gave the answer. Since both methods gave me 64, and the equality condition is satisfied, I think that 64 is indeed the minimum.

But just to be thorough, let me test with some numbers. Suppose I take a = b = 1/8, c = 1/4, d = 1/2. Then the sum is 1/8 + 1/8 + 1/4 + 1/2 = 1/8 + 1/8 + 2/8 + 4/8 = 8/8 = 1. Correct. And the expression is 8 + 8 + 16 + 32 = 64. Now, let me try another set of numbers to see if I can get a lower value.

Suppose I take a = 0.2, b = 0.2, c = 0.2, d = 0.4. Then the sum is 0.2 + 0.2 + 0.2 + 0.4 = 1. Then compute the expression:

1/0.2 + 1/0.2 + 4/0.2 + 16/0.4 = 5 + 5 + 20 + 40 = 70, which is higher than 64. So worse.

Another test: suppose I take a = 0.1, b = 0.1, c = 0.2, d = 0.6. Then sum is 0.1 + 0.1 + 0.2 + 0.6 = 1. The expression is 1/0.1 + 1/0.1 + 4/0.2 + 16/0.6 = 10 + 10 + 20 + 26.666... ≈ 66.666..., which is still higher than 64.

Another test: suppose I take a = 1/8, b = 1/8, c = 1/4, d = 1/2. Which gives 64. What if I take a slightly different value? Let me try a = 0.12, b = 0.12, then c = 2a = 0.24, d = 4a = 0.48. Then sum is 0.12 + 0.12 + 0.24 + 0.48 = 0.96. Wait, that's less than 1. So need to adjust. Let me make a = 0.125, which is 1/8, then same as before. If I take a = 0.1, then c = 0.2, d = 0.4, and then b = ?

Wait, if I take a = 0.1, then since a = b, b = 0.1. Then c = 0.2, d = 0.4. Then total sum is 0.1 + 0.1 + 0.2 + 0.4 = 0.8. So need to add 0.2 more. Maybe adjust variables. Suppose I take a = 0.1 + x, b = 0.1 + x, c = 0.2 + y, d = 0.4 + z, such that (0.1 + x) + (0.1 + x) + (0.2 + y) + (0.4 + z) = 1. Which simplifies to 0.8 + 2x + y + z = 1 ⇒ 2x + y + z = 0.2. But this might complicate. Maybe not the best approach.

Alternatively, take a = 0.1, b = 0.1, c = 0.2, d = 0.6. Wait, that was tried before. The value was higher. So maybe 64 is indeed the minimum.

Alternatively, let me try to use AM-GM on the terms. Let me think. Suppose I consider the terms $\frac{1}{a}$, $\frac{1}{b}$, $\frac{4}{c}$, $\frac{16}{d}$. Let me think of them as four terms. Let me try to apply AM-GM on these four terms. The AM-GM inequality says that the arithmetic mean is greater or equal to the geometric mean. So:

$$
\frac{\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}}{4} \ge \sqrt[4]{\frac{1}{a} \cdot \frac{1}{b} \cdot \frac{4}{c} \cdot \frac{16}{d}} = \sqrt[4]{\frac{64}{abcd}}
$$

Therefore,

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge 4 \sqrt[4]{\frac{64}{abcd}} = 4 \cdot \left( \frac{64}{abcd} \right)^{1/4}
$$

But this seems not directly helpful because we don't know the value of abcd. However, maybe we can relate abcd to the constraint a + b + c + d = 1. But with AM-GM on a + b + c + d, we have:

$$
\frac{a + b + c + d}{4} \ge \sqrt[4]{abcd} \Rightarrow \frac{1}{4} \ge \sqrt[4]{abcd} \Rightarrow abcd \le \left( \frac{1}{4} \right)^4 = \frac{1}{256}
$$

Therefore, substituting back into the previous inequality:

$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge 4 \cdot \left( \frac{64}{abcd} \right)^{1/4} \ge 4 \cdot \left( \frac{64}{1/256} \right)^{1/4} = 4 \cdot \left( 64 \times 256 \right)^{1/4}
$$

Calculating 64 × 256: 64 × 256 = 16384. Then, 16384^(1/4). Let me compute that. 16384 is 2^14, since 2^10 = 1024, 2^14 = 16384. Therefore, 2^14^(1/4) = 2^(14/4) = 2^(3.5) = 2^3 * 2^0.5 = 8 * sqrt(2) ≈ 11.3137. Therefore, the lower bound would be 4 * 11.3137 ≈ 45.25. But this is way lower than 64, which suggests that this approach is not tight. Therefore, the AM-GM approach here is not giving a useful bound because the equality conditions are not met. Therefore, this method isn't helpful here. So, the previous methods of Cauchy-Schwarz and Lagrange multipliers are better.

Alternatively, maybe I can use the method of substitution. Let me try to express variables in terms of a single variable. Since from the Lagrange multipliers, we found that a = b, c = 2a, d = 4a. Therefore, the variables are dependent. Therefore, the minimal value occurs when variables are in this proportion. Therefore, the minimal value is 64.

Alternatively, maybe I can think of the problem as optimizing with weights. Let me think of the expression as $\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}$. Let me consider variables x = a, y = b, z = c, w = d. Then, the problem is to minimize $\frac{1}{x} + \frac{1}{y} + \frac{4}{z} + \frac{16}{w}$ with x + y + z + w = 1.

Alternatively, maybe I can use the method of substitution by letting variables be proportional to the coefficients in the denominators. Wait, in the Lagrange multipliers method, we found that the ratios between variables are determined by the coefficients in the expression. For example, since the partial derivatives are proportional to the coefficients in the denominators.

Alternatively, think of the problem as minimizing the sum of terms with different weights. Maybe using the concept of convexity. Since the function is convex, the minimum is achieved at the critical point found by Lagrange multipliers.

Alternatively, since both methods gave me 64, and testing with other values gives higher results, I think that 64 is indeed the correct minimum. Therefore, the answer is 64.

But just to make sure, let me try another set of variables. Suppose I take a = 1/8, b = 1/8, c = 1/4, d = 1/2. Then, if I slightly perturb one variable, say increase a by ε, then decrease another variable to keep the sum 1. Let me try:

Let me take a = 1/8 + ε, b = 1/8 - ε, c = 1/4, d = 1/2. Then, the sum is (1/8 + ε) + (1/8 - ε) + 1/4 + 1/2 = 1/8 + 1/8 + 1/4 + 1/2 = same as before, 1. Now compute the expression:

1/a + 1/b + 4/c + 16/d.

So 1/(1/8 + ε) + 1/(1/8 - ε) + 4/(1/4) + 16/(1/2).

Compute the first two terms:

Let me approximate for small ε. Let me take ε approaching 0.

1/(1/8 + ε) ≈ 8 - 64ε (using the expansion 1/(a + ε) ≈ 1/a - ε/a² for small ε)

Similarly, 1/(1/8 - ε) ≈ 8 + 64ε.

Adding them together: (8 - 64ε) + (8 + 64ε) = 16. So the first two terms sum to 16, same as before. The other terms are 16 + 32 = 48. So total is 16 + 48 = 64. Therefore, even after perturbation, the value remains the same? Wait, that's strange. Wait, but if I take ε ≠ 0, then the first two terms would be 1/(1/8 + ε) + 1/(1/8 - ε). Let me compute exactly:

Let me compute 1/(1/8 + ε) + 1/(1/8 - ε) = [ (1/8 - ε) + (1/8 + ε) ] / [ (1/8 + ε)(1/8 - ε) ] = (2/8) / [ (1/64 - ε²) ] = (1/4) / (1/64 - ε²) = (1/4) * [1 / (1/64 - ε²)].

Which is equal to (1/4) * [ 64 / (1 - 64 ε²) ] = 16 / (1 - 64 ε²) ≈ 16(1 + 64 ε²) for small ε. Therefore, the first two terms are approximately 16 + 16 * 64 ε². Therefore, the total expression is approximately 16 + 16 * 64 ε² + 48 = 64 + 1024 ε², which is greater than 64. Therefore, perturbing a and b slightly increases the value. Therefore, this suggests that the critical point is indeed a minimum.

Similarly, if I perturb other variables, say, increase c and decrease d, keeping the sum same. Let me try. Let me take c = 1/4 + ε, d = 1/2 - ε. Then, the sum remains 1. Compute the expression:

1/a + 1/b + 4/c + 16/d. Since a and b are still 1/8, so 8 + 8 = 16. Then 4/c = 4/(1/4 + ε) ≈ 16 - 64 ε (using similar expansion). And 16/d = 16/(1/2 - ε) ≈ 32 + 128 ε. Therefore, total expression is 16 + (16 - 64 ε) + (32 + 128 ε) = 16 + 16 + 32 + ( -64 ε + 128 ε ) = 64 + 64 ε. Therefore, if ε is positive, the value increases, and if ε is negative, the value decreases? Wait, but if ε is negative, then c = 1/4 + ε would be less than 1/4, and d = 1/2 - ε would be more than 1/2. Let me check with ε negative. Let me take ε = -k, where k > 0. Then:

4/c = 4/(1/4 - k) ≈ 16 + 64k (since 1/(1/4 - k) ≈ 4/(1 - 4k) ≈ 4(1 + 4k) = 4 + 16k, so 4/(1/4 - k) ≈ 16 + 64k? Wait, maybe my previous expansion was wrong. Let me recast:

Let me compute 4/c where c = 1/4 + ε. Let me write c = 1/4 + ε. Then 4/c = 4 / (1/4 + ε) = 16 / (1 + 4ε) ≈ 16(1 - 4ε + 16ε² - ...) ≈ 16 - 64ε + ... Similarly, 16/d where d = 1/2 - ε. Then 16/d = 32 / (1 - 2ε) ≈ 32(1 + 2ε + 4ε² + ...) ≈ 32 + 64ε + ... Therefore, adding them together: (16 - 64ε) + (32 + 64ε) = 48. Therefore, the total expression is 16 + 48 = 64. Wait, so if I take ε positive or negative, the first order terms cancel out? Wait, but earlier when I considered the expansion, I thought there was a term with ε. Maybe my previous calculation was wrong. Let me check again.

Wait, if I take c = 1/4 + ε, then 4/c = 4 / (1/4 + ε) = 16 / (1 + 4ε) ≈ 16(1 - 4ε + 16ε² - ...) ≈ 16 - 64ε + ... Similarly, d = 1/2 - ε, so 16/d = 16 / (1/2 - ε) = 32 / (1 - 2ε) ≈ 32(1 + 2ε + 4ε² + ...) ≈ 32 + 64ε + ... Therefore, adding these two terms: (16 - 64ε) + (32 + 64ε) = 48. Therefore, the first order terms cancel. Therefore, the total expression is 16 + 48 = 64, same as before. Therefore, the second order terms would be 16*(16ε²) + 32*(4ε²) = 256ε² + 128ε² = 384ε². Therefore, the total expression is 64 + 384ε², which is greater than 64. Therefore, perturbing c and d also leads to an increase in the value. Therefore, this suggests that the critical point is indeed a minimum.

Therefore, all these tests confirm that 64 is the minimum value. Therefore, I can be confident that the answer is 64.

**Final Answer**
\boxed{64}

To find the minimum value of the expression
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}
$$
under the constraint
$$
a + b + c + d = 1
$$
with $ a, b, c, d > 0 $, we can use the **Cauchy-Schwarz inequality** in a strategic way.

---

### Step 1: Apply Cauchy-Schwarz Inequality

We rewrite the expression as:
$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d}
$$

Now apply the **Cauchy-Schwarz inequality** in the form:
$$
\left( \sum \frac{x_i^2}{a_i} \right) \ge \frac{(\sum x_i)^2}{\sum a_i}
$$

Let:
- $ x_1 = 1, x_2 = 1, x_3 = 2, x_4 = 4 $
- $ a_1 = a, a_2 = b, a_3 = c, a_4 = d $

Then:
$$
\frac{1^2}{a} + \frac{1^2}{b} + \frac{2^2}{c} + \frac{4^2}{d} \ge \frac{(1 + 1 + 2 + 4)^2}{a + b + c + d}
$$

Compute:
- Numerator: $ (1 + 1 + 2 + 4)^2 = 8^2 = 64 $
- Denominator: $ a + b + c + d = 1 $

So:
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \ge \frac{64}{1} = 64
$$

---

### Step 2: Equality Condition

Equality in Cauchy-Schwarz holds when:
$$
\frac{x_1}{a_1} = \frac{x_2}{a_2} = \frac{x_3}{a_3} = \frac{x_4}{a_4}
$$

That is:
$$
\frac{1}{a} = \frac{1}{b} = \frac{2}{c} = \frac{4}{d}
$$

From this, we deduce:
- $ a = b $
- $ c = 2a $
- $ d = 4a $

Substitute into the constraint:
$$
a + b + c + d = a + a + 2a + 4a = 8a = 1 \Rightarrow a = \frac{1}{8}
$$

Then:
- $ a = b = \frac{1}{8} $
- $ c = \frac{1}{4} $
- $ d = \frac{1}{2} $

---

### Step 3: Compute the Expression

Now compute:
$$
\frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d}
= \frac{1}{1/8} + \frac{1}{1/8} + \frac{4}{1/4} + \frac{16}{1/2}
= 8 + 8 + 16 + 32 = 64
$$

---

### Step 4: Confirm Minimum

We have shown that the expression is **greater than or equal to 64** by Cauchy-Schwarz, and that the value **64 is achievable** when the variables are set as above. Therefore, the **minimum value** of the expression is:

$$
\boxed{64}
$$

Solution:
By Cauchy-Schwarz,
\[(a + b + c + d) \left( \frac{1}{a} + \frac{1}{b} + \frac{4}{c} + \frac{16}{d} \right) \ge (1 + 1 + 2 + 4)^2 = 64.\]Equality occurs when $a = b = \frac{c}{2} = \frac{d}{4}$ and $a + b + c + d = 1.$  We can solve to get $a = \frac{1}{8},$ $b = \frac{1}{8},$ $c = \frac{1}{4},$ and $d = \frac{1}{2},$ so the minimum value is $\boxed{64}.$