# What is the unit vector that is orthogonal to the plane containing  (2i + 3j – 7k)  and  (3i – 4j + 4k) ?

Apr 10, 2017

The unit vector is =〈-16/sqrt1386,-29/sqrt1386,-17/sqrt1386〉

#### 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=〈2,3,-7〉 and vecb=〈3,-4,4〉

Therefore,

$| \left(\vec{i} , \vec{j} , \vec{k}\right) , \left(2 , 3 , - 7\right) , \left(3 , - 4 , 4\right) |$

$= \vec{i} | \left(3 , - 7\right) , \left(- 4 , 4\right) | - \vec{j} | \left(2 , - 7\right) , \left(3 , 4\right) | + \vec{k} | \left(2 , 3\right) , \left(3 , - 4\right) |$

$= \vec{i} \left(3 \cdot 4 - 7 \cdot 4\right) - \vec{j} \left(2 \cdot 4 + 7 \cdot 3\right) + \vec{k} \left(- 2 \cdot 4 - 3 \cdot 3\right)$

=〈-16,-29,-17〉=vecc

Verification by doing 2 dot products

〈-16,-29,-17〉.〈2,3,-7〉=-16*2-29*3-7*17=0

〈-16,-29,-17〉.〈3,-4,4〉=-16*3+29*4-17*4=0

So,

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

The unit vector is

=vecc/||vecc||=1/sqrt(16^2+29^2+17^2)〈-16,-29,-17〉

=1/sqrt1386〈-16,-29,-17〉