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

Dec 19, 2016

The answer is =〈0,-3/sqrt13,-2/sqrt13〉

#### Explanation:

We do a cross product to find the vector orthogonal to the plane

The vector is given by the determinant

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

$= \hat{i} \left(6 - 6\right) - \hat{j} \left(9 - - 3\right) + \hat{k} \left(- 6 - 2\right)$

=〈0,-12,-8〉

Verification by doing the dot product

〈0,-12,-8〉.〈3,2,-3〉=0-24+24=0

〈0,-12,-8〉.〈1,-2,3〉=0+24-24=0

The vector is orthgonal to the other 2 vectors

The unit vector is obtained by dividing by the modulus

∥〈0,-12,-8〉∥=sqrt(0+144+64)=sqrt208=4sqrt13

Thre unit vector is =1/(4sqrt13)〈0,-12,-8〉

=〈0,-3/sqrt13,-2/sqrt13〉