# What is the cross product of [4, -4, 4] and [-6, 5, 1] ?

Jul 19, 2018

\begin{pmatrix}-24&-28&-4\end{pmatrix}

#### Explanation:

Use the following cross product formula:

(u1,u2,u3)xx(v1,v2,v3) = (u2v3 − u3v2 , u3v1 − u1v3 , u1v2 − u2v1)

(4,-4,4)xx(-6,5,1) = (-4*1 − 4*5 , 4*-6 − 4*1 , 4*5 − -4*-6)

$= \left(- 24 , - 28 , - 4\right)$

Jul 19, 2018

The vector is = 〈-24,-28,-4〉

#### Explanation:

The cross product of 2 vectors is calculated with the determinant

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(d , e , f\right) , \left(g , h , i\right) |$

where veca=〈d,e,f〉 and vecb=〈g,h,i〉 are the 2 vectors

Here, we have veca=〈4,-4,4〉 and vecb=〈-6,5,1〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(4 , - 4 , 4\right) , \left(- 6 , 5 , 1\right) |$

$= \vec{i} | \left(- 4 , 4\right) , \left(5 , 1\right) | - \vec{j} | \left(4 , 4\right) , \left(- 6 , 1\right) | + \vec{k} | \left(4 , - 4\right) , \left(- 6 , 5\right) |$

$= \vec{i} \left(\left(- 4\right) \cdot \left(1\right) - \left(5\right) \cdot \left(4\right)\right) - \vec{j} \left(\left(4\right) \cdot \left(1\right) - \left(- 6\right) \cdot \left(4\right)\right) + \vec{k} \left(\left(4\right) \cdot \left(5\right) - \left(- 4\right) \cdot \left(- 6\right)\right)$

=〈-24,-28,-4〉=vecc

Verification by doing 2 dot products

〈4,-4,4〉.〈-24,-28,-4〉=(4)*(-24)+(-4)*(-28)+(4)*(-4)=0

〈-24,-28,-4〉.〈-6,5,1〉=(-24)*(-6)+(-28)*(5)+(-4)*(1)=0

So,

$\vec{c}$ is perpendicular to $\vec{a}$ and $\vec{b}$