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

Jan 27, 2016

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)

= $\left(\left(- 2 \cdot 1\right) - \left(3 \cdot - 1\right)\right) i + \left(\left(3 \cdot 1\right) - \left(1 \cdot 1\right)\right) j + \left(\left(1 \cdot - 1\right) - \left(- 2 \cdot 1\right)\right) k$

= $\left(- 2 - \left(- 3\right)\right) i + \left(3 - 1\right) j + \left(- 1 - \left(- 2\right)\right) k$

= $\left(i + 2 j + k\right)$

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 $\left(a i + b j + c k\right)$ is given by:

$l = \sqrt{{a}^{2} + {b}^{2} + {c}^{2}}$

In this case:

$l = \sqrt{{1}^{2} + {2}^{2} + {1}^{2}} = \sqrt{6}$

Dividing each component of $\left(i + 2 j + k\right)$ by $\sqrt{6}$ yields our answer, which is that the unit vector orthogonal to the plane in which (i−2j+3k) and (i−j+k) lie is:

$\left(\frac{i}{\sqrt{6}} + \frac{2}{\sqrt{6}} j + \frac{k}{\sqrt{6}}\right)$