Register to reply

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

by skyturnred
Tags: hint, product, proof, simple, solving, vector
Share this thread:
skyturnred
#1
Feb12-12, 05:50 PM
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:

[itex]\frac{1}{||v||^{2}}[/itex][itex]\overline{v}[/itex]=[itex]\frac{1}{||u||^{2}}[/itex][itex]\overline{u}[/itex]
Phys.Org News Partner Science news on Phys.org
FIXD tells car drivers via smartphone what is wrong
Team pioneers strategy for creating new materials
Team defines new biodiversity metric
LCKurtz
#2
Feb12-12, 08:57 PM
HW Helper
Thanks
PF Gold
LCKurtz's Avatar
P: 7,659
Quote Quote by skyturnred View Post
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:

[itex]\frac{1}{||v||^{2}}[/itex][itex]\overline{v}[/itex]=[itex]\frac{1}{||u||^{2}}[/itex][itex]\overline{u}[/itex]
Think about what the identity $$
\vec A\cdot \vec B = |\vec A||\vec B|\cos\theta$$implies about this problem.
skyturnred
#3
Feb14-12, 01:35 PM
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.

LCKurtz
#4
Feb14-12, 02:00 PM
HW Helper
Thanks
PF Gold
LCKurtz's Avatar
P: 7,659
Please give me a hint to solving this simple vector dot product proof

Quote Quote by skyturnred View Post
(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 ...
skyturnred
#5
Feb14-12, 02:18 PM
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?
LCKurtz
#6
Feb14-12, 02:22 PM
HW Helper
Thanks
PF Gold
LCKurtz's Avatar
P: 7,659
Quote Quote by skyturnred View Post
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 Quote by LCKurtz View Post
Unless ##\cos\theta=0##, in which case ...
Quote Quote by skyturnred View Post
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##?
skyturnred
#7
Feb14-12, 02:26 PM
P: 116
Quote Quote by LCKurtz View Post
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!


Register to reply

Related Discussions
Please give me some hint Calculus & Beyond Homework 2
Vector Algebra - Vector Triple Product Proof Calculus & Beyond Homework 14
Someone give me a hint.. been at it for hrs... Advanced Physics Homework 11
Give hint to this problem... Precalculus Mathematics Homework 4
Give me please hint General Math 1