# What is the unit vector that is orthogonal to the plane containing  (29i-35j-17k)  and  (32i-38j-12k) ?

Jan 18, 2017

The answer is =1/299.7〈-226,-196,18〉

#### Explanation:

The vector perpendiculatr 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=〈29,-35,-17〉 and vecb=〈32,-38,-12〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(29 , - 35 , - 17\right) , \left(32 , - 38 , - 12\right) |$

$= \vec{i} | \left(- 35 , - 17\right) , \left(- 38 , - 12\right) | - \vec{j} | \left(29 , - 17\right) , \left(32 , - 12\right) | + \vec{k} | \left(29 , - 35\right) , \left(32 , - 38\right) |$

$= \vec{i} \left(35 \cdot 12 - 17 \cdot 38\right) - \vec{j} \left(- 29 \cdot 12 + 17 \cdot 32\right) + \vec{k} \left(- 29 \cdot 38 + 35 \cdot 32\right)$

=〈-226,-196,18〉=vecc

Verification by doing 2 dot products

〈-226,-196,18〉.〈29,-35,-17〉=-226*29+196*35-17*18=0

〈-226,-196,18〉.〈32,-38,-12〉=-226*32+196*38-12*18=0

So,

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

The unit vector is

=1/sqrt(226^2+196^2+18^2)〈-226,-196,18〉

=1/299.7〈-226,-196,18〉