# What is the unit vector that is normal to the plane containing (- 4i + 5 j-k) and (2i+ j - 3k)?

Nov 10, 2017

The unit vector is $= < - \frac{1}{\sqrt{3}} , - \frac{1}{\sqrt{3}} , - \frac{1}{\sqrt{3}} >$

#### Explanation:

The normal vector perpendicular to a plane is calculated with the determinant

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

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

Here, we have veca=〈-4,5,-1〉 and vecb=〈2,1,-3〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(- 4 , 5 , - 1\right) , \left(2 , 1 , - 3\right) |$

$= \vec{i} | \left(5 , - 1\right) , \left(1 , - 3\right) | - \vec{j} | \left(- 4 , - 1\right) , \left(2 , - 3\right) | + \vec{k} | \left(- 4 , 5\right) , \left(2 , 1\right) |$

$= \vec{i} \left(5 \cdot - 3 + 1 \cdot 1\right) - \vec{j} \left(4 \cdot 3 + 1 \cdot 2\right) + \vec{k} \left(- 4 \cdot 1 - 2 \cdot 5\right)$

=〈-14,-14,-14〉=vecc

Verification by doing 2 dot products

〈-14,-14,-14〉.〈-4,5,-1〉=-14*-4+-14*5+14*1=0

〈-14,-14,-14〉.〈2,1,-3〉=-28-14+14*3=0

So,

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

$| | \vec{c} | | = \sqrt{{14}^{2} + {14}^{2} + {14}^{2}} = 14 \sqrt{3}$

The unit vector is

hatc=1/(||vecc||)vecc=1/(14sqrt3)〈-14,-14,-14〉#

$= < - \frac{1}{\sqrt{3}} , - \frac{1}{\sqrt{3}} , - \frac{1}{\sqrt{3}} >$