# What is the unit vector that is orthogonal to the plane containing  ( - 4 i - 5 j + 2 k)  and  (4 i + 4 j + 2 k) ?

May 13, 2018

The unit vector is 1/sqrt(596)*〈-18,16,4〉

#### Explanation:

A vector that is orthogonal to $2$ other vectors is calculated with the cross product. The latter is calculate 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,-5,2〉 and vecb=〈4,4,2〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(- 4 , - 5 , 2\right) , \left(4 , 4 , 2\right) |$

$= \vec{i} | \left(- 5 , 2\right) , \left(4 , 2\right) | - \vec{j} | \left(- 4 , 2\right) , \left(4 , 2\right) | + \vec{k} | \left(- 4 , - 5\right) , \left(4 , 4\right) |$

$= \vec{i} \left(\left(- 5\right) \cdot \left(2\right) - \left(4\right) \cdot \left(2\right)\right) - \vec{j} \left(\left(- 4\right) \cdot \left(2\right) - \left(4\right) \cdot \left(2\right)\right) + \vec{k} \left(\left(- 4\right) \cdot \left(4\right) - \left(- 5\right) \cdot \left(4\right)\right)$

=〈-18,16,4〉=vecc

Verification by doing 2 dot products

〈-18,16,4〉.〈-4,-5,2〉=(-18)*(-4)+(16)*(-5)+(4)*(2)=0

〈-18,16,4〉.〈4,4,2〉=(-18)*(4)+(16)*(4)+(4)*(2)=0

So,

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

The unit vector is

$\hat{c} = \frac{\vec{c}}{| | \vec{c} | |}$

The magnitude of $\vec{c}$ is

||vecc||=||〈-18,16,4〉||=sqrt((-18)^2+(16)^2+(4)^2)

$= \sqrt{596}$

The unit vector is 1/sqrt(596)*〈-18,16,4〉