# What is the unit vector that is orthogonal to the plane containing  ( i - 2 j + 3 k)  and  (4 i + 4 j + 2 k) ?

Feb 22, 2016

There are two steps in solving this question: (1) taking the cross product of the vectors and then (2) normalizing the resultant. In this case, the final unit vector is $\left(- \frac{16}{\sqrt{500}} i + \frac{10}{\sqrt{500}} j + \frac{12}{\sqrt{500}} k\right)$ or $\left(- \frac{16}{22.4} i + \frac{10}{22.4} j + \frac{12}{22.4} k\right)$.

#### Explanation:

First step: cross product of the vectors.

(i-2j+3k)xx(4i+4j+2k) = (((-2)*2-3*4))i+(3*4-1*2)j+(1*4-(-2)*4)k)=((-4-12)i+(12-2)j+(4-(-8))k)=(-16i+10j+12k)

Second step: normalize the resultant vector.

To normalize a vector we divide each element by the length of the vector. To find the length:

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

Putting it all together, the unit vector orthogonal to the given vectors can be represented as:

$\left(- \frac{16}{\sqrt{500}} i + \frac{10}{\sqrt{500}} j + \frac{12}{\sqrt{500}} k\right)$ or $\left(- \frac{16}{22.4} i + \frac{10}{22.4} j + \frac{12}{22.4} k\right)$