# Please give me a hint to solving this simple vector dot product proof

by skyturnred
Tags: hint, product, proof, simple, solving, vector
 P: 116 1. The problem statement, all variables and given/known data Let 'u' and 'v' be two non zero vectors such that the prjection of 'u' along 'v' equals the projection of 'v' along 'u.' Using the formula for projection, show that 'u' and 'v' are either perpendicular or parallel. 2. Relevant equations 3. The attempt at a solution Please don't just answer it, I would like to do this one on my own. But I first need a hint because I have been trying for about 30 minutes. I know that the projection of 'u' along 'v' is u dot v, divided by the square of the norm of 'v'. Then this scalar is multiplied through 'v'. But that's about all I have. Edit: I guess I said more about how far I got. I get the following: $\frac{1}{||v||^{2}}$$\overline{v}$=$\frac{1}{||u||^{2}}$$\overline{u}$
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PF Gold
P: 7,583
 Quote by skyturnred 1. The problem statement, all variables and given/known data Let 'u' and 'v' be two non zero vectors such that the prjection of 'u' along 'v' equals the projection of 'v' along 'u.' Using the formula for projection, show that 'u' and 'v' are either perpendicular or parallel. Please don't just answer it, I would like to do this one on my own. But I first need a hint because I have been trying for about 30 minutes. I know that the projection of 'u' along 'v' is u dot v, divided by the square of the norm of 'v'. Then this scalar is multiplied through 'v'. But that's about all I have. Edit: I guess I said more about how far I got. I get the following: $\frac{1}{||v||^{2}}$$\overline{v}$=$\frac{1}{||u||^{2}}$$\overline{u}$
Think about what the identity $$\vec A\cdot \vec B = |\vec A||\vec B|\cos\theta$$implies about this problem.
 P: 116 (sorry for the long time before replying) Thanks! That hint definitely helped me out! In case anyone comes by this thread seeking the same as I did, when you apply the identity that LCKurtz showed us, you get a scalar number multiplied by vector v equals vector u. Obviously, this means they are either parallel or anti-parallel.
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PF Gold
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Please give me a hint to solving this simple vector dot product proof

 Quote by skyturnred (sorry for the long time before replying) Thanks! That hint definitely helped me out! In case anyone comes by this thread seeking the same as I did, when you apply the identity that LCKurtz showed us, you get a scalar number multiplied by vector v equals vector u. Obviously, this means they are either parallel or anti-parallel.
Unless ##\cos\theta=0##, in which case ...
 P: 116 But then wouldn't that just mean that they are parallel anyways? So no matter how you look at it, if proj u over v equals proj v over u, as long as these aren't zero vectors, wouldn't the angle HAVE to be 0 or 180?
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Thanks
PF Gold
P: 7,583
 Quote by skyturnred 1. The problem statement, all variables and given/known data Let 'u' and 'v' be two non zero vectors such that the prjection of 'u' along 'v' equals the projection of 'v' along 'u.' Using the formula for projection, show that 'u' and 'v' are either perpendicular or parallel.
 Quote by LCKurtz Unless ##\cos\theta=0##, in which case ...
 Quote by skyturnred But then wouldn't that just mean that they are parallel anyways? So no matter how you look at it, if proj u over v equals proj v over u, as long as these aren't zero vectors, wouldn't the angle HAVE to be 0 or 180?
What do you get for ##\vec i \cdot \vec j##?
P: 116
 Quote by LCKurtz What do you get for ##\vec i \cdot \vec j##?
Oh I see.. so then given no extra information about u and v, they both can be EITHER parallel OR perpendicular? I guess I misread the question.. I thought it was asking to show that it was one OR the other, and because of this, I blindly looked for one until I found it, and then stopped before I checked the other.

Definitely one vector dotted into another equals 0 if they are both perpendicular.

Thanks again!

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