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From Anton, I have this simple looking LA question:

Find all unit vectors parallel to the yz plane that are perpendicular to the vector 3,1,-2

Since this vector is sloped in all 3 dimensions, and since yz is I flat in the x dimension, I am baffled how this can have a solution.

I found a near exact version of this question on another forum but the vector given did have a zero component.

I'm clearly missing some obvious detail, a quick point out would be much appreciated.

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Any vector in the $i$ direction is perpendicular to the yz plane. So you need a vector perpendicular both to $i$ and the vector $<3,1,-2>$. How do you find a vector perpendicular to two given vectors?

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  • $\begingroup$ You cross them. 'i direction' is more or less synonymous with 'x direction', yes? Thanks for the reply. $\endgroup$ – GregT Jun 9 '15 at 1:22
  • $\begingroup$ @GregT, yes, that's correct. Sorry for the delayed reply. $\endgroup$ – Paul Jun 10 '15 at 12:22
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friend ,the above scenario will have an answer, initially i thought what you are thinking,but the real case is that you can actually draw a perpendicular to both of i direction as well as the vector ,ok think it this way a vector has been drawn from origin to any arbitrary point then you draw a perpendicular from the vector perpendicular to the x axis(in y direction) , and now you get a perpendicular to i as well as yz plane.Therefore it has a solution!!CHEERS!

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