# How do you find two unit vectors orthogonal to both i-j+k and 4j+4k?

Mar 5, 2017

$\pm \frac{1}{\sqrt{6}} \left(- 2 , - 1 , 1\right) .$

#### Explanation:

Let us recall that, for vectors $\vec{a} \mathmr{and} \vec{b}$, their Vector

(Cross) Product , i.e., $\vec{a} \times \vec{b}$ is Orthogonal to both

$\vec{a} \mathmr{and} \vec{b} .$

The Desired Unit Vectors, then, can be obtained by,

$\pm \frac{\vec{a} \times \vec{b}}{|} | \left(\vec{a} \times \vec{b}\right) | |$.

Now, with veca=i-j+k=(1,-1,1), &, vecb=4j+4k=4(0,1,1), we have,

$\vec{a} \times \vec{b} = | \left(i , j , k\right) , \left(1 , - 1 , 1\right) , \left(0 , 4 , 4\right) | = 4 | \left(i , j , k\right) , \left(1 , - 1 , 1\right) , \left(0 , 1 , 1\right) |$

$= 4 \left(- 2 , - 1 , 1\right)$,

$\Rightarrow | | \left(\vec{a} \times \vec{b}\right) | | = 4 \sqrt{{\left(- 2\right)}^{2} + {\left(- 1\right)}^{1} + {1}^{2}} = 4 \sqrt{6.}$

Hence, the desired vectors are, $\frac{\pm 4 \left(- 2 , - 1 , 1\right)}{4 \sqrt{6}} , \mathmr{and} ,$

$\pm \frac{1}{\sqrt{6}} \left(- 2 , - 1 , 1\right) .$

Enjoy Maths.!