# What is the unit vector that is orthogonal to the plane containing  (8i + 12j + 14k)  and  (2i+j+2k) ?

Aug 6, 2016

Two steps are required:

1. Take the cross product of the two vectors.
2. Normalise that resultant vector to make it a unit vector (length of 1).

The unit vector, then, is given by:

$\left(\frac{10}{\sqrt{500}} i + \frac{12}{\sqrt{500}} j - \frac{16}{\sqrt{500}} k\right)$

#### Explanation:

1. The cross product is given by:

$\left(8 i + 12 j + 14 k\right) \times \left(2 i + j + 2 k\right)$
$= \left(\left(12 \cdot 2 - 14 \cdot 1\right) i + \left(14 \cdot 2 - 8 \cdot 2\right) j + \left(8 \cdot 1 - 12 \cdot 2\right) k\right)$
$= \left(10 i + 12 j - 16 k\right)$

1. To normalise a vector, find its length and divide each coefficient by that length.

$r = \sqrt{{10}^{2} + {12}^{2} + {\left(- 16\right)}^{2}} = \sqrt{500} \approx 22.4$

The unit vector, then, is given by:

$\left(\frac{10}{\sqrt{500}} i + \frac{12}{\sqrt{500}} j - \frac{16}{\sqrt{500}} k\right)$