## Problem Statement
There are an infinite number of vectors **v** such that:
$$
\left(\begin{array}{c}
1 \\
2 \\
-5
\end{array}\right) \times \mathbf{v} = \left(\begin{array}{c}
90 \\
30 \\
30
\end{array}\right)
$$
Find the vector **v** that has the smallest magnitude.
---
## Response
To find the vector **v** that satisfies the equation:
$$
\left(\begin{array}{c}
1 \\
2 \\
-5
\end{array}\right) \times \mathbf{v} = \left(\begin{array}{c}
90 \\
30 \\
30
\end{array}\right)
$$
and has the smallest magnitude, we start by letting:
$$
\mathbf{v} = \left(\begin{array}{c}
x \\
y \\
z
\end{array}\right)
$$
The cross product:
$$
\left(\begin{array}{c}
1 \\
2 \\
-5
\end{array}\right) \times \left(\begin{array}{c}
x \\
y \\
z
\end{array}\right)
$$
is given by:
$$
\left(\begin{array}{c}
2z + 5y \\
-5z + x \\
2x - y
\end{array}\right)
$$
We set this equal to:
$$
\left(\begin{array}{c}
90 \\
30 \\
30
\end{array}\right)
$$
giving us the system of equations:
$$
\begin{cases}
2z + 5y = 90 \\
-5z + x = 30 \\
2x - y = 30
\end{cases}
$$
---
## Solving the System
### Step 1: Solve for $ y $
From the first equation:
$$
2z + 5y = 90 \implies y = \frac{90 - 2z}{5}
$$
### Step 2: Substitute $ y $ into the third equation
Substitute $ y = \frac{90 - 2z}{5} $ into $ 2x - y = 30 $:
$$
2x - \frac{90 - 2z}{5} = 30 \implies 10x - (90 - 2z) = 150 \implies 10x + 2z = 240 \implies 5x + z = 120
$$
### Step 3: Solve for $ x $ and $ z $
From $ 5x + z = 120 $, we get:
$$
z = 120 - 5x
$$
Substitute $ z = 120 - 5x $ into $ -5z + x = 30 $:
$$
-5(120 - 5x) + x = 30 \implies -600 + 25x + x = 30 \implies 26x = 630 \implies x = \frac{630}{26} = \frac{315}{13}
$$
Then:
$$
z = 120 - 5\left(\frac{315}{13}\right) = \frac{1560 - 1575}{13} = \frac{-15}{13}
$$
Substitute $ x = \frac{315}{13} $ and $ z = \frac{-15}{13} $ into $ y = \frac{90 - 2z}{5} $:
$$
y = \frac{90 - 2\left(\frac{-15}{13}\right)}{5} = \frac{90 + \frac{30}{13}}{5} = \frac{1170 + 30}{65} = \frac{1200}{65} = \frac{240}{13}
$$
---
## Final Vector
The vector **v** with the smallest magnitude is:
$$
\mathbf{v} = \left(\begin{array}{c}
\frac{315}{13} \\
\frac{240}{13} \\
\frac{-15}{13}
\end{array}\right)
$$
---
## Verification
Compute the cross product:
$$
\left(\begin{array}{c}
1 \\
2 \\
-5
\end{array}\right) \times \left(\begin{array}{c}
\frac{315}{13} \\
\frac{240}{13} \\
\frac{-15}{13}
\end{array}\right)
$$
Result:
$$
\left(\begin{array}{c}
2\left(\frac{-15}{13}\right) + 5\left(\frac{240}{13}\right) \\
-5\left(\frac{-15}{13}\right) + \frac{315}{13} \\
2\left(\frac{315}{13}\right) - \frac{240}{13}
\end{array}\right)
= \left(\begin{array}{c}
\frac{-30 + 1200}{13} \\
\frac{75 + 315}{13} \\
\frac{630 - 240}{13}
\end{array}\right)
= \left(\begin{array}{c}
\frac{1170}{13} \\
\frac{390}{13} \\
\frac{390}{13}
\end{array}\right)
= \left(\begin{array}{c}
90 \\
30 \\
30
\end{array}\right)
$$
---
## Conclusion
The vector **v** that satisfies the equation and has the smallest magnitude is:
$$
\boxed{\left(\begin{array}{c}
\frac{315}{13} \\
\frac{240}{13} \\
\frac{-15}{13}
\end{array}\right)}
$$
