# How do you find a unit vectors are orthogonal to both i+j and i+k?

Nov 19, 2016

$\pm \left(- \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}}\right)$

#### Explanation:

a=i+j=<1, 1, 0) and b=i+k=<1, 0, 1>

Let $c = + < \left(\cos \alpha , \cos \beta , \cos \gamma\right) >$ be the unit vectors (in

opposite directions) orthogonal to $a \mathmr{and} b$.

Then the scalar product $c . a = \cos \alpha + \cos \beta = 0$.

Similarly, $c . b = \cos \alpha + \cos \gamma = 0$.

It follows that $c = \pm < - \cos \alpha , \cos \alpha , \cos \alpha >$.

The directions are equally inclined to the axes, in the respective

octant ( the 2nd OX'YZ and 8th OXY'Z') , and so,

$\cos \alpha = \pm \frac{1}{\sqrt{3}}$

The answer is $\pm \left(- \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}} , \frac{1}{\sqrt{3}}\right)$