Question

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

Answer 1

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

`| (i,j,k), (u_1,u_2,u_3), (v_1,v_2,v_3) |`

`vecw=| (i,j,k), (2,3,-7), (3,-1,-2) |`

`=i(-6-7)-j(-4+21)+k(-2-9)`

`=i(-13)+j(-17)+k(-11)`

`=〈-13,-17,-11〉`

To verify that `vecw` is perpendicular to `vecu` and `vecv`

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`, `vecw` is perpendicular to `vecu` and `vecv`

To calculate the unit vector, we divide by the modulus

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