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

Nov 17, 2016

The vector is =〈-14,-18,-15〉

#### Explanation:

Let vecu=〈3,1,-4〉 and vecv=〈3,-4,2〉

The cross product is given by the determinant

$\vec{u}$ x $\vec{v}$ $= | \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(3 , 1 , - 4\right) , \left(3 , - 4 , 2\right) |$

$= \vec{i} | \left(1 , - 4\right) , \left(- 4 , 2\right) | - \vec{j} | \left(3 , - 4\right) , \left(3 , 2\right) | + \vec{k} | \left(3 , 1\right) , \left(3 , - 4\right) |$

$= \vec{i} \left(2 - 16\right) + \vec{j} \left(- 6 - 12\right) + \vec{k} \left(- 12 - 3\right)$

=vecw=〈-14,-18,-15〉

Verification, the dot products must de $0$

vecu.vecw=〈3,1,-4〉.〈-14,-18,-15〉=(-42-18+60)=0

vecv.vecw=〈3,-4,2〉.〈-14,-18,-15〉=(-42+72-30)=0

Therefore, $\vec{w}$ is perpendicular to $\vec{u}$ and $\vec{v}$