How do you find a unit vector perpendicular to both vector u(1, -1,-1)and vector v(2, -2, 3)?

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Answer 1 · 2016-07-11T07:44:33

$+-(1/sqrt 2, 1/sqrt 2, 0)$

The vector perpendicular to both $u and v$ is $+-(uXv)/|uXv|$

$=+-((1. -1, -1)X(2, -2, 3))/|(1. -1, -1)X(2, -2, 3)|$

Now, the numerator vector is

$((-1)(3)-(-1)(-2), (-1)(2)-(1)(3), (1)(-2)-(-1)(2))$

$=+-(-5, -5, 0)$ and $|(-5, -5, 0)|=5 sqrt 2$ '

So, the answer is $+-$ (-5, -5, 0) $/(5 sqrt 2)$

$=+-(1/sqrt 2, 1/sqrt 2, 0)$ .