# What is the unit vector that is normal to the plane containing (- 3 i + j -k) and (i+2j+2k)?

Mar 25, 2017

The answer is $= < \frac{4}{\sqrt{90}} , \frac{5}{\sqrt{90}} , - \frac{7}{\sqrt{90}} >$

#### Explanation:

The vector perpendicular to 2 vectors is calculated with the determinant (cross product)

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

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

Here, we have veca=〈-3,1,-1〉 and vecb=〈1,2,2〉

Therefore,

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

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

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

=〈4,5,-7〉=vecc

Verification by doing 2 dot products

〈4,5,-7〉.〈-3,1,-1〉=-12+5+7=0

〈4,5,-7〉.〈1,2,2〉=4+10-14=0#

So,

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

The unit vector is

$= \frac{1}{\sqrt{16 + 25 + 49}} \cdot < 4 , 5 , - 7 >$

$= < \frac{4}{\sqrt{90}} , \frac{5}{\sqrt{90}} , - \frac{7}{\sqrt{90}} >$