# How do you find a unit vector parallel to the xz plane and perpendicular to i - 2j + 3k?

Aug 27, 2016

$\pm \frac{1}{\sqrt{10}} < 3 , 0 , - 1 >$

#### Explanation:

Unit vector in the direction making angles

$\alpha , \beta \mathmr{and} \gamma$ with the positive x, y and z

axes is .

$< \cos \alpha , \cos \beta , \cos \gamma >$.

If the vector is parallel to xz-plane, $\beta$ is a right angle.

So, $\cos \beta = 0$.

And so, the required vector is $< \cos \alpha , 0 , \cos \gamma >$ that is

perpendicular to $< 1 , - 2 , 3 >$.

#< cos alpha, 0, cos gamma>.<1, -2, 3>=0, giving

$\cos \alpha + 3 \cos \gamma = 0$, and so, $\cos \gamma = - \frac{1}{3} \cos \alpha$. .

Thus, any such vector is $\cos \alpha < 1 , 0 , - \frac{1}{3} >$, and for unit vector,

$\cos \alpha = \pm \frac{3}{\sqrt{10}}$.

The answer is $\pm \frac{1}{\sqrt{10}} < 3 , 0. - 1 >$.