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

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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>#.