<Question>
There are an infinite number of vectors $\mathbf{v}$ such that
\[\begin{pmatrix} 1 \\ 2 \\ -5 \end{pmatrix} \times \mathbf{v} = \begin{pmatrix} 90 \\ 30 \\ 30 \end{pmatrix}.\]Find the vector $\mathbf{v}$ that has the smallest magnitude.
</Question>

<think>
<Plan_1>
Okay, so I need to find the vector v such that when I take the cross product of (1, 2, -5) with v, I get (90, 30, 30). And among all possible such vectors, I need to find the one with the smallest magnitude. Hmm, let's start by recalling some vector algebra concepts.

First, the cross product of two vectors is another vector that's perpendicular to both of the original vectors. The magnitude of the cross product is related to the area of the parallelogram formed by the vectors, but maybe that's not directly useful here.

The cross product is defined as:

<prompt_1.1> Recall the definition of the cross product and its properties. </prompt_1.1>

The cross product in three dimensions can be computed using the determinant formula. If we have vectors a = (a1, a2, a3) and b = (b1, b2, b3), then their cross product a × b is:

<prompt_1.2> Write the formula for the cross product of two vectors in three dimensions. </prompt_1.2>

So, given that (1, 2, -5) × v = (90, 30, 30), I need to solve for v. Let me denote v = (v1, v2, v3). Then the cross product (1, 2, -5) × v is:

<prompt_1.3> Set up the equation for the cross product (1, 2, -5) × v and equate it to (90, 30, 30). </prompt_1.3>
</Plan_1>

<Execution_1>
<execution_1.1>
If I have two vectors a and b, then a × b = ||a|| ||b|| sin(theta) n, where theta is the angle between them and n is the unit vector perpendicular to both. So, the result is a vector, not a scalar. The cross product is only defined in three dimensions, and it's anti-commutative, meaning b × a = - (a × b).
</execution_1.1>

<execution_1.2>
a × b = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1).

So, given that (1, 2, -5) × v = (90, 30, 30), I need to solve for v. Let me denote v = (v1, v2, v3). Then the cross product (1, 2, -5) × v is:

First component: (2*v3 - (-5)*v2) = 2v3 + 5v2

Second component: (-5*v1 - 1*v3) = -5v1 - v3

Third component: (1*v2 - 2*v1) = v2 - 2v1

And this should equal (90, 30, 30).
</execution_1.2>

<execution_1.3>
So, writing out the determinant:

i  j  k
1  2 -5
v1 v2 v3

Calculating this determinant, we get:

i*(2*v3 - (-5)*v2) - j*(1*v3 - (-5)*v1) + k*(1*v2 - 2*v1)

Simplifying each component:

- The i component: 2v3 + 5v2
- The j component: -(v3 + 5v1)
- The k component: v2 - 2v1

So, the cross product is (2v3 + 5v2, -(v3 + 5v1), v2 - 2v1). And this is equal to (90, 30, 30).
</execution_1.3>
</Execution_1>

<Plan_2>
Based on execution_1.3:
Therefore, we can set up the following system of equations:

<prompt_2.1> Set up the system of equations from the cross product components. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
1. 2v3 + 5v2 = 90

2. -(v3 + 5v1) = 30 → -v3 -5v1 = 30 → 5v1 + v3 = -30

3. v2 - 2v1 = 30

So, three equations with three variables: v1, v2, v3.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let me write them again:

1. 2v3 + 5v2 = 90

2. 5v1 + v3 = -30

3. v2 - 2v1 = 30

Hmm, so equation 3 can be solved for v2: v2 = 30 + 2v1

Then equation 2: 5v1 + v3 = -30 → v3 = -30 -5v1

Then substitute these into equation 1:

2v3 +5v2 = 90

Substituting v3 and v2:

<prompt_3.1> Solve the system of equations using substitution. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
2*(-30 -5v1) +5*(30 +2v1) = 90

Let me compute this step by step:

First term: 2*(-30) + 2*(-5v1) = -60 -10v1

Second term: 5*30 +5*2v1 = 150 +10v1

Adding both terms together: (-60 -10v1) + (150 +10v1) = (-60 +150) + (-10v1 +10v1) = 90 + 0 =90

So, 90 =90. Hmm, that's an identity, which means that the equations are dependent, and there's a free variable. So, the system has infinitely many solutions depending on v1. Therefore, the general solution is:

v2 =30 +2v1

v3= -30 -5v1

So, vector v can be written as:

v = [v1, 30 +2v1, -30 -5v1]

Which can be rewritten as:

v = v1*[1, 2, -5] + [0,30, -30]

So, the solutions are all vectors of the form [1,2,-5] scaled by v1 plus the particular solution [0,30,-30].
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
So, to find the vector v with the smallest magnitude, we need to minimize ||v||^2, which is v1^2 + v2^2 + v3^2. Since v2 =30 +2v1 and v3= -30 -5v1, substitute these into the magnitude squared:

<prompt_4.1> Set up the expression for minimizing the magnitude of v. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
||v||^2 = v1^2 + (30 +2v1)^2 + (-30 -5v1)^2

Let me compute each term:

First term: v1^2

Second term: (30 +2v1)^2 = 30^2 + 2*30*2v1 + (2v1)^2 = 900 +120v1 +4v1^2

Third term: (-30 -5v1)^2 = (-30)^2 + 2*(-30)*(-5v1) + (-5v1)^2 = 900 +300v1 +25v1^2

So, adding all three terms:

v1^2 + 900 +120v1 +4v1^2 +900 +300v1 +25v1^2

Combine like terms:

v1^2 +4v1^2 +25v1^2 = 30v1^2

120v1 +300v1 = 420v1

900 +900 =1800

So, ||v||^2 =30v1^2 +420v1 +1800
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Now, to find the minimum of this quadratic function in v1. Since the coefficient of v1^2 is positive (30), the parabola opens upwards, so the minimum occurs at the vertex. The vertex of a quadratic ax² +bx +c is at x = -b/(2a).

<prompt_5.1> Find the value of v1 that minimizes the magnitude of v using calculus. </prompt_5.1>
</Plan_5>

<Execution_5>
<execution_5.1>
Here, a =30, b=420.

So, v1 = -420/(2*30) = -420/60 = -7

So, the value of v1 that minimizes the magnitude is -7.

Therefore, substituting back into v2 and v3:

v2 =30 +2*(-7) =30 -14=16

v3= -30 -5*(-7)= -30 +35=5

Thus, the vector v is ( -7, 16, 5 )
</execution_5.1>
</Execution_5>

<Plan_6>
Based on execution_5.1:
Wait, but let me check if this is correct. Let me verify the cross product.

<prompt_6.1> Verify the solution by checking if the cross product is correct. </prompt_6.1>

But let me make sure that this is indeed the minimal solution. Since we derived the quadratic in v1 and found the minimum, this should be the minimal magnitude.

Alternatively, another approach could be to recognize that the general solution is of the form v = particular solution + homogeneous solution. The homogeneous solution is any vector in the null space of the cross product operator, i.e., vectors v such that (1,2,-5) × v =0. These are the vectors parallel to (1,2,-5). So, the general solution is the particular solution plus a scalar multiple of (1,2,-5). To find the vector with minimal magnitude, we can subtract the component of the particular solution in the direction of the null space vector. That is, using vector projection.

<prompt_6.2> Verify the solution using the alternative method of vector projection. </prompt_6.2>

Let me compute ||v||^2 for v=(-7,16,5):

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

Wait, but let me check if there is a possibility of a miscalculation here. Let me recheck the cross product calculation.

<prompt_6.3> Recheck the cross product calculation and confirm the minimal magnitude. </prompt_6.3>

But let me also check if there is another way to approach this problem, perhaps using the inverse of the cross product? But cross product isn't invertible in the same way as scalar multiplication because it's not injective. However, if the vectors are not parallel, we can find a unique solution, but in this case, since the cross product equation has infinitely many solutions, we have to find the one with minimal norm.

<prompt_6.4> Explore other methods and confirm the solution. </prompt_6.4>

But let me check once more if there is a possible mistake in the algebra. Let's retrace the steps.

<prompt_6.5> Perform a final verification of the solution. </prompt_6.5>
</Plan_6>

<Execution_6>
<execution_6.1>
Original vectors are (1,2,-5) and v=(-7,16,5). Let's compute their cross product:

Using the determinant formula:

i  j  k
1  2 -5
-7 16 5

Calculating determinant:

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

Compute each component:

i: 2*5 =10; (-5)*16 = -80; so 10 - (-80) =90 → i*90

j: -( [1*5 =5] - [(-5)*(-7)=35] ) = -(5 -35) = -(-30) =30 → j*30

k:1*16=16; 2*(-7)=-14; so 16 - (-14)=30 → k*30

So, the cross product is (90,30,30), which matches the given vector.
</execution_6.1>

<execution_6.2>
Wait, maybe I should have considered that approach first. Let me see.

Alternatively, the minimal vector would be the particular solution minus the projection onto the null space. Wait, but in our case, the null space is the set of vectors parallel to (1,2,-5). So, if the general solution is v = v_p + t*(1,2,-5), where v_p is a particular solution, then the vector with minimal magnitude is obtained by choosing t such that v_p - t*(1,2,-5) is orthogonal to the null space direction. Wait, actually, the minimal norm solution in this case would be the particular solution minus the projection of the particular solution onto the null space. Wait, maybe I need to think in terms of orthogonal decomposition.

Alternatively, since the problem is to find the vector v with the smallest magnitude such that (1,2,-5) × v = b (where b is (90,30,30)), then the minimal solution is the particular solution minus the component in the null space. The null space here is the set of vectors parallel to (1,2,-5). So, if we write the general solution as v = v_p + t*(1,2,-5), then to minimize ||v||, we can set the derivative to zero with respect to t, which is equivalent to projecting v_p onto the null space and subtracting it. Wait, actually, the minimal norm solution is given by v = v_p - proj_n(v_p), where n is the null space direction. But since the null space is one-dimensional, spanned by (1,2,-5), the projection of v_p onto the null space is (v_p ⋅ n / ||n||²) * n.

Let me compute that. Let me take v_p as the particular solution we found earlier: v_p = (-7,16,5). Wait, but actually, how did we get v_p? We solved the system and set v1 = -7. But perhaps another approach is to find a particular solution first. Let me choose t=0 in the general solution. Wait, but in our previous approach, we found the general solution as v = [1,2,-5]v1 + [0,30,-30]. So, when v1 = -7, we get v = (-7,16,5). But perhaps another way is to find a particular solution by choosing a specific value for the free variable. Alternatively, maybe there's a different particular solution. Wait, but in any case, the minimal solution is unique, and our calculation gives (-7,16,5).
</execution_6.2>

<execution_6.3>
Wait, the cross product (1,2,-5) × (-7,16,5) should be (90,30,30). Let's compute it:

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

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

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

Yes, that's correct. So, the cross product is indeed (90,30,30). So, the vector (-7,16,5) is a solution.

Now, the question is whether this is the vector with the smallest magnitude. Since we found it by minimizing the quadratic expression, which is the standard method for such problems, this should be the minimal one.
</execution_6.3>

<execution_6.4>
Wait, but in our case, the cross product equation (1,2,-5) × v = (90,30,30). We found that the solutions are of the form v = [1,2,-5] * v1 + [0,30,-30]. Then, to find the vector with minimal magnitude, we set up the quadratic in v1 and found the minimum at v1=-7, leading to v=(-7,16,5). Let's verify this cross product.

Compute (1,2,-5) × (-7,16,5):

Using the cross product formula:

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

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

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

Yes, that gives (90,30,30), which matches the required cross product. So, this vector is indeed a solution.

Now, is this the vector with the smallest magnitude? Since we derived it by minimizing the quadratic form, which is the standard way to find the minimal norm solution in such underdetermined systems (infinitely many solutions), this should be the correct answer.
</execution_6.4>

<execution_6.5>
Original equations:

1. 2v3 +5v2=90

2. 5v1 +v3= -30

3. v2 -2v1=30

Expressed v2=30 +2v1, v3= -30 -5v1. Substituted into first equation:

2*(-30 -5v1) +5*(30 +2v1)= -60 -10v1 +150 +10v1=90. So, yes, that's an identity, so the system is consistent and has infinitely many solutions.

Then, expressing ||v||^2=30v1^2 +420v1 +1800. Then, vertex at v1=-420/(2*30)= -7. Then, plugging back gives v=(-7,16,5). Then, the cross product (1,2,-5) × (-7,16,5).

Compute cross product:

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

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

Wait, hold on, cross product formula: The j component is negative of (a1v3 - a3v1). Wait, let's compute it properly.

Cross product formula:

Given a = (a1,a2,a3), b=(b1,b2,b3)

a × b = (a2b3 -a3b2, a3b1 -a1b3, a1b2 -a2b1)

So, here a=(1,2,-5), b=(-7,16,5)

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

Second component: (-5)*(-7) -1*5=35 -5=30. But in the cross product formula, the second component is a3b1 -a1b3, which is (-5)*(-7) -1*5=35 -5=30. However, in the cross product formula, the second component is subtracted, so the cross product is (90, -(30), 30). Wait, no, wait:

Wait, no. Wait, the cross product components are:

First component: a2b3 -a3b2

Second component: a3b1 -a1b3

Third component: a1b2 -a2b1

So, the second component is a3b1 -a1b3. So, in this case, a3 is -5, b1 is -7, so -5*(-7)=35. a1 is 1, b3 is 5, so 1*5=5. Therefore, 35 -5=30. So, the second component is 30. However, in the cross product formula, the second component is subtracted in the determinant formula. Wait, no, in the determinant, the j component is multiplied by -1. Let me clarify:

The cross product determinant is:

i * (a2b3 -a3b2) - j * (a1b3 -a3b1) + k * (a1b2 -a2b1)

Therefore, the second component (j component) is negative of (a1b3 -a3b1). So, in our case, the second component is -[(1)(5) - (-5)(-7)] = -[5 -35] = -[-30] =30. Which matches. So, the cross product is (90,30,30), which is exactly the vector we wanted. Therefore, the solution is correct.

Therefore, the minimal magnitude vector is (-7,16,5). Its magnitude squared is 49 +256 +25=330, so magnitude is sqrt(330). Since 330 is 10*33=10*3*11, so sqrt(330) is simplest form.

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}}.
</execution_6.5>
</Execution_6>

<Plan_7>
Based on execution_6.5:
**Final Answer**
The vector \(\mathbf{v}\) with the smallest magnitude is \boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}.
</Plan_7>

<Execution_7>
No prompt to execute.
</Execution_7>

<Final_answer>
Therefore, the vector with the smallest magnitude that satisfies the given condition is $\boxed{\begin{pmatrix} -7 \\ 16 \\ 5 \end{pmatrix}}$.
</Final_answer>
</think>