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

Question

8115 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-27T07:39:47

$+- 1/sqrt 10<3, 0,-1>$

Unit vector in the direction making angles

$alpha, beta and gamma$ 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, $cos gamma =-1/3 cos alpha$ . .

Thus, any such vector is $cos alpha<1, 0, -1/3>$ , and for unit vector,

$cos alpha =+-3/sqrt 10$ .

The answer is $+- 1/sqrt 10<3, 0. -1>$ .