# What is the unit vector that is orthogonal to the plane containing  (20j +31k)  and  (32i-38j-12k) ?

Nov 13, 2017

The unit vector is $= = \frac{1}{1507.8} < 938 , 992 , - 640 >$

#### Explanation:

The vector orthogonal to 2 vectros in 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

Here, we have veca=〈0,20,31〉 and vecb=〈32,-38,-12〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(0 , 20 , 31\right) , \left(32 , - 38 , - 12\right) |$

$= \vec{i} | \left(20 , 31\right) , \left(- 38 , - 12\right) | - \vec{j} | \left(0 , 31\right) , \left(32 , - 12\right) | + \vec{k} | \left(0 , 20\right) , \left(32 , - 38\right) |$

$= \vec{i} \left(20 \cdot - 12 + 38 \cdot 31\right) - \vec{j} \left(0 \cdot - 12 - 31 \cdot 32\right) + \vec{k} \left(0 \cdot - 38 - 32 \cdot 20\right)$

=〈938,992,-640〉=vecc

Verification by doing 2 dot products

〈938,992,-640〉.〈0,20,31〉=938*0+992*20-640*31=0

〈938,992,-640〉.〈32,-38,-12〉=938*32-992*38+640*12=0

So,

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

The unit vector is

$\hat{c} = \frac{\vec{c}}{|} | \vec{c} | | = \frac{< 938 , 992 , - 640 >}{|} | < 938 , 992 , - 640 > | |$

$= \frac{1}{1507.8} < 938 , 992 , - 640 >$