Vector Projection: Is it Possible for projuv=projvu?

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Homework Statement



Is it possible for

projuv=projvu


Homework Equations





The Attempt at a Solution



This can only occur if:

[tex]\frac{|\mathbf{u\cdot v}|}{^{\|u\|^{2}}}\mathbf{u} = \frac{|\mathbf{u\cdot v}|}{^{\|v\|^{2}}}\mathbf{v}[/tex]

So if either is the zero vector, it is true. How can I prove that it can only be true if either is the zero vector, or v=u?
 
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You can immediately see from

[tex] \frac{|\mathbf{u\cdot v}|}{^{\|u\|^{2}}}\mathbf{u} = \frac{|\mathbf{u\cdot v}|}{^{\|v\|^{2}}}\mathbf{v}[/tex]

That u is a multiple of v (and vice versa). What do you think beyond that?
 
Office_Shredder said:
You can immediately see from



That u is a multiple of v (and vice versa). What do you think beyond that?

That if one is larger/smaller than the other, the projections cannot be equivalent. Is that sufficient to prove it?
 
IniquiTrance said:
That if one is larger/smaller than the other, the projections cannot be equivalent.

Do you have a reason for believing that?

If u is parallel to v, then

[tex] \frac{|\mathbf{u\cdot v}|}{^{\|u\|^{2}}}\mathbf{u}[/tex]

is parallel to
[tex]\frac{|\mathbf{u\cdot v}|}{^{\|v\|^{2}}}\mathbf{v}[/tex]

So how can we tell whether they are equal?
 
Is it because:

[tex]\frac{\mathbf{u}}{^{\|u\|}}\frac{1}{\|u\|} = \frac{\mathbf{v}}{^{\|v\|}}\frac{1}{\|v\|}[/tex]

So we have a constant times a unit vector on each side, therefore the constants must be equal?