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

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

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Answer 1 · 2016-07-21T07:15:32

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.

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

1 Answer

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