Question

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

Answer 1

General solution is `(+-(2+c, 3 c, -3))/|((2+c, 3 c, -3))|`
Particular solutions for + sign, and c =0 and 1, are
`1/sqrt 13(2, 0, -3) and 1/sqrt 3(1, 1, -1)`, respectively.

Explanation:

A vector parallel to yz-plane is `p=j+ck=(0, 1, c)`, with c at your

choice..

The other given vector is `q=3 i-j+2 k=(3, -1, 2)`

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

`=+-(2+c, 3 c, -3)` is

orthogonal to both `p and q`..

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

`1/sqrt 13(2, 0, -3) and 1/sqrt 3(1, 1, -1)`, respectively.

The general solution is

`(+-(2+c, 3 c, -3))/|((2+c, 3 c, -3))|`, where c i a parameter.