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

Nov 30, 2016

The answer is =1/sqrt579*〈-13,-17,-11〉

#### Explanation:

To calculate a vector perpendicular to two other vectors, you have to calculate the cross product

Let vecu=〈2,3,-7〉 and vecv=〈3,-1,-2〉

The cross product is given by the determinant

$| \left(i , j , k\right) , \left({u}_{1} , {u}_{2} , {u}_{3}\right) , \left({v}_{1} , {v}_{2} , {v}_{3}\right) |$

$\vec{w} = | \left(i , j , k\right) , \left(2 , 3 , - 7\right) , \left(3 , - 1 , - 2\right) |$

$= i \left(- 6 - 7\right) - j \left(- 4 + 21\right) + k \left(- 2 - 9\right)$

$= i \left(- 13\right) + j \left(- 17\right) + k \left(- 11\right)$

=〈-13,-17,-11〉

To verify that $\vec{w}$ is perpendicular to $\vec{u}$ and $\vec{v}$

We do a dot product.

vecw.vecu=〈-13,-17,-11〉.〈2,3,-7〉=-26--51+77=0

vecw.vecv=〈-13,-17,-11〉.〈3,-1,-2〉=-39+17+22=0

As the dot products $= 0$, $\vec{w}$ is perpendicular to $\vec{u}$ and $\vec{v}$

To calculate the unit vector, we divide by the modulus

hatw=vecw/(∥vecw∥)=1/sqrt579*〈-13,-17,-11〉