# What is the cross product of <-3,0, 2> and <-1, -4, 1>?

Jul 27, 2018

$< 8 , 1 , 12 >$

#### Explanation:

We make a 3x3 matrix, and find the determinant of it to find the cross product of the two vectors.

$\left(\begin{matrix}i & j & k \\ - 3 & 0 & 2 \\ - 1 & - 4 & 1\end{matrix}\right)$

The determinant of this matrix is:

$i \cdot \det \left(\begin{matrix}0 & 2 \\ - 4 & 1\end{matrix}\right) - j \cdot \det \left(\begin{matrix}- 3 & 2 \\ - 1 & 1\end{matrix}\right) + k \cdot \det \left(\begin{matrix}- 3 & 0 \\ - 1 & - 4\end{matrix}\right)$

$i \left[\left(0\right) \left(1\right) - \left(2\right) \left(- 4\right)\right] - j \left[\left(- 3\right) \left(1\right) - \left(2\right) \left(- 1\right)\right] + k \left[\left(- 3\right) \left(- 4\right) - \left(0\right) \left(- 1\right)\right]$

$i \left(0 + 8\right) - j \left(- 3 + 2\right) + k \left(12 + 0\right)$

$i \left(8\right) - j \left(- 1\right) + k \left(12\right)$

$8 i + j + 12 k$

In other words, the cross product of $< - 3 , 0 , 2 >$ and $< - 1 , - 4 , 1 >$ is the vector $< 8 , 1 , 12 >$.