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how do i find all unit vectors orthogonal to v=<3,4,0>
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u.v=0... all vectors
\[u _{1}+3=0\]\[u _{1}=-3\]
err... wait... hmm....
\[3u _{1}+4u _{2}=0\]
we need another equation....
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u1² + u2² = 1
what pops to mind is u cross v divided by magnitude u cross v
okay... so given your vector... there is no z... so the slope in the floor is y/x or 4/3 a perpendicular slope would be -3/4 which leads to <4,-3,0> or <-4,3,0> as perpendicular vectors in the floor. to make them unit vectors in the floor we would divide by the magnitude to get\[\frac{1}{5}<4,-3,0> or \frac{1}{5}<-4,3,0>\]in the floor.
|dw:1347772027565:dw|
|dw:1347772166980:dw|now, how do we get a z component....
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\[\frac{1}{\sqrt{25+z ^{2}}}<4,-3,z> or \frac{1}{\sqrt{25+z ^{2}}}<-4,3,z>\]
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