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

Apr 6, 2017

The unit vector is =〈2/sqrt150,-5/sqrt150,-11/sqrt150〉

#### 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=〈-4,5,-3〉

Therefore,

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

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

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

=〈2,-5,-11〉=vecc

Verification by doing 2 dot products

〈2,-5,-11〉.〈-3,1,-1〉=-6-5+11=0

〈2,-5,-11〉.〈-4,5,-3〉=-8-25+33=0

So,

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

The unit vector is

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

=1/sqrt(4+25+121)〈2,-5,-11〉

=1/sqrt150〈2,-5,-11〉#