# What is the cross product of [3, 1, -4] and [2, 6, -1] ?

Aug 24, 2016

$= 23 \hat{x} - 5 \hat{y} + 16 \hat{z}$

#### Explanation:

the cross product you seek is the determinant of the following matrix

$\left(\begin{matrix}\hat{x} & \hat{y} & \hat{z} \\ 3 & 1 & - 4 \\ 2 & 6 & - 1\end{matrix}\right)$

$= \hat{x} \left(1 \cdot \left(- 1\right) - \left(- 4\right) \cdot 6\right) - \hat{y} \left(3 \setminus \cdot \left(- 1\right) - \left(- 4\right) \cdot 2\right) + \hat{z} \left(3 \cdot 6 - 2 \cdot 1\right)$

$= 23 \hat{x} - 5 \hat{y} + 16 \hat{z}$

this should be perpendicular to these 2 vectors and we can check that via the scalar dot product

$< 23 , - 5 , 16 > \cdot < 3 , 1 , - 4 > = 69 - 5 - 64 = 0$

$< 23 , - 5 , 16 > \cdot < 2 , 6 , - 1 > = 46 - 30 - 16 = 0$