# How do you find two unit vectors which are orthogonal to the vector 3i-j+2k and parallel to the yz plane?

Jul 21, 2016

General solution is $\frac{\pm \left(2 + c , 3 c , - 3\right)}{|} \left(\left(2 + c , 3 c , - 3\right)\right) |$
Particular solutions for + sign, and c =0 and 1, are
$\frac{1}{\sqrt{13}} \left(2 , 0 , - 3\right) \mathmr{and} \frac{1}{\sqrt{3}} \left(1 , 1 , - 1\right)$, respectively.

#### Explanation:

A vector parallel to yz-plane is $p = j + c k = \left(0 , 1 , c\right)$, with c at your

choice..

The other given vector is $q = 3 i - j + 2 k = \left(3 , - 1 , 2\right)$

Now, r=+-(p X q)=+-((0, 1, c) X (3, -1, 2)

$= \pm \left(2 + c , 3 c , - 3\right)$ is

orthogonal to both $p \mathmr{and} q$..

Such orthogonal unit vectors, for + sign, and c =0 and 1, are

$\frac{1}{\sqrt{13}} \left(2 , 0 , - 3\right) \mathmr{and} \frac{1}{\sqrt{3}} \left(1 , 1 , - 1\right)$, respectively.

The general solution is

$\frac{\pm \left(2 + c , 3 c , - 3\right)}{|} \left(\left(2 + c , 3 c , - 3\right)\right) |$, where c i a parameter.