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

Jun 11, 2018

The unit vector is =1/sqrt(2870)〈17,-30,-41〉

#### Explanation:

First calculate the vector orthogonal to the other $2$ vectors. This is given by the cross product.

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

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

Here, we have veca=〈-4,-5,2〉 and vecb=〈-5,4,-5〉

Therefore,

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

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

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

=〈17,-30,-41〉=vecc

Verification by doing 2 dot products

〈17,-30,-41〉.〈-4,-5,2〉=(17)*(-4)+(-30)*(-5)+(-41)*(2)=0

〈17,-30,-41〉.〈-5,4,-5〉=(17)*(-5)+(-30)*(4)+(-41)*(-5)=0

So,

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

The unit vector is

hatc=vecc/(||vecc||)=1/sqrt(17^2+(-30)^2+(-41)^2)*〈17,-30,-41〉

=1/sqrt(2870)〈17,-30,-41〉