Question

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

Answer 1

There are two steps in finding this solution: 1. Find the cross product of the two vectors to find a vector orthogonal to the plane containing them and 2. normalise that vector so that it has unit length.

Explanation:

The first step in solving this problem is to find the cross product of the two vectors. The cross product by definition finds a vector orthogonal to the plane in which the two vectors being multiplied lie.

`(i−2j+3k) xx (i−j+k)`

= `((-2*1)-(3*-1))i+((3*1)-(1*1))j+((1*-1)-(-2*1))k`

= `(-2-(-3))i+(3-1)j+(-1-(-2))k`

= `(i+2j+k)`

This is a vector orthogonal to the plane, but it is not yet a unit vector. To make it one we need to 'normalise' the vector: divide each of its components by its length. The length of a vector `(ai+bj+ck)` is given by:

`l = sqrt(a^2+b^2+c^2)`

In this case:

`l = sqrt(1^2+2^2+1^2) = sqrt6`

Dividing each component of `(i+2j+k)` by `sqrt6` yields our answer, which is that the unit vector orthogonal to the plane in which `(i−2j+3k) and (i−j+k)` lie is:

`(i/sqrt6+2/sqrt6j+k/sqrt6)`