Projection Vectors Problem

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February 25th 2013, 03:18 PM
anonymoususer20

Projection Vectors Problem

The angle between two vectors a and b is ∅, where ∅≠90°. Under what conditions will |Proj_ab|² + |Proj_ba|² = 1?

Don't really know where to start on this one.
February 25th 2013, 03:44 PM
ILikeSerena

Re: Projection Vectors Problem

Quote:

Originally Posted by anonymoususer20

The angle between two vectors a and b is ∅, where ∅≠90°. Under what conditions will |Proj_ab|² + |Proj_ba|² = 1?

Don't really know where to start on this one.

Hi anonymoususer20! :)

I'm not entirely sure what you mean with Proj_ab.
Presumably it is:

$\text{Proj}_{ab} = \mathbf a \cdot \hat{\mathbf b}$

where $\hat{\mathbf b}$ is the unit vector in the direction of $\mathbf b$ .

If that is the case you can substitute $\mathbf a \cdot \hat{\mathbf b} = a \cos \phi$ .
And you can substitute something similar for the other projection.

Have you tried that?
If so, what did you get?
February 25th 2013, 04:17 PM
anonymoususer20

Re: Projection Vectors Problem

Proj_ab is a projection vector.

|P_ab|= |a.b|/|b|.
February 25th 2013, 04:21 PM
ILikeSerena

Re: Projection Vectors Problem

Quote:

Originally Posted by anonymoususer20

Proj_ab is a projection vector.

|P_ab|= |a.b|/|b|.

Aha! Then it is exactly what I suspected, since $\mathbf{\hat b} = \frac{\mathbf b}{|\mathbf b|} = \frac{\mathbf b}{b}$ .
February 25th 2013, 06:58 PM
Plato

Re: Projection Vectors Problem

Quote:

Originally Posted by anonymoususer20

Proj_ab is a projection vector.

|P_ab|= |a.b|/|b|.

Quote:

Originally Posted by ILikeSerena

Aha! Then it is exactly what I suspected, since $\mathbf{\hat b} = \frac{\mathbf b}{|\mathbf b|} = \frac{\mathbf b}{b}$ .

There seems to be somewhat of confusion on notation here.
I have never seen the notation $\text{Proj}_{ab}$ .
In North American text books $\text{Proj}_{\vec{b}}\;\vec{a}=\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b}{$

If we use that definition then $\left\|\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b}\right\|^2=| \vec{a}\cdot\vec{b}|^2$

Now that makes a really interesting question.
February 25th 2013, 10:41 PM
ILikeSerena

Re: Projection Vectors Problem

Quote:

Originally Posted by Plato

There seems to be somewhat of confusion on notation here.
I have never seen the notation $\text{Proj}_{ab}$ .
In North American text books $\text{Proj}_{\vec{b}}\;\vec{a}=\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b}{$

If we use that definition then $\left\|\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b}\right\|^2=| \vec{a}\cdot\vec{b}|^2$

Now that makes a really interesting question.

That does make more sense and is probably what it means.
I didn't think to include the direction.

However, $\left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 = \frac{|\vec{a}\cdot\vec{b}|^2}{|\vec{b}|^2} = (a \cos \phi)^2$
February 26th 2013, 04:03 AM
Plato

Re: Projection Vectors Problem

Quote:

Originally Posted by ILikeSerena

That does make more sense and is probably what it means.
I didn't think to include the direction.
However, $\left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 = \frac{|\vec{a}\cdot\vec{b}|^2}{|\vec{b}|^2} = (a \cos \phi)^2$

Actually no, it equals $\left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 =|{\vec{a}\cdot \vec{b}|^2$ .

So we get $|{\vec{a}\cdot \vec{b}|^2=\frac{1}{2}$
February 26th 2013, 04:11 AM
ILikeSerena

Re: Projection Vectors Problem

$\left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 = \frac{|\vec a \cdot \vec b|^2} {|\vec{b}\cdot\vec{b}|^2} ||\vec b||^2 = \frac{|\vec a \cdot \vec b|^2} {||\vec{b}||^4} ||\vec b||^2 = \frac{|\vec a \cdot \vec b|^2} {||\vec{b}||^2}$