# Projection Vectors Problem

• Feb 25th 2013, 03:18 PM
anonymoususer20
Projection Vectors Problem
The angle between two vectors a and b is ∅, where ∅≠90°. Under what conditions will |Projab|2 + |Projba|2 = 1?

Don't really know where to start on this one.
• Feb 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 |Projab|2 + |Projba|2 = 1?

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

Hi anonymoususer20! :)

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

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

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

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

Have you tried that?
If so, what did you get?
• Feb 25th 2013, 04:17 PM
anonymoususer20
Re: Projection Vectors Problem
Projab is a projection vector.

|Pab|= |a.b|/|b|.
• Feb 25th 2013, 04:21 PM
ILikeSerena
Re: Projection Vectors Problem
Quote:

Originally Posted by anonymoususer20
Projab is a projection vector.

|Pab|= |a.b|/|b|.

Aha! Then it is exactly what I suspected, since $\displaystyle \mathbf{\hat b} = \frac{\mathbf b}{|\mathbf b|} = \frac{\mathbf b}{b}$.
• Feb 25th 2013, 06:58 PM
Plato
Re: Projection Vectors Problem
Quote:

Originally Posted by anonymoususer20
Projab is a projection vector.

|Pab|= |a.b|/|b|.

Quote:

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

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

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

Now that makes a really interesting question.
• Feb 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 $\displaystyle \text{Proj}_{ab}$.
In North American text books $\displaystyle \text{Proj}_{\vec{b}}\;\vec{a}=\frac{\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b}{$

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

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, $\displaystyle \left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 = \frac{|\vec{a}\cdot\vec{b}|^2}{|\vec{b}|^2} = (a \cos \phi)^2$
• Feb 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, $\displaystyle \left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 = \frac{|\vec{a}\cdot\vec{b}|^2}{|\vec{b}|^2} = (a \cos \phi)^2$

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

So we get $\displaystyle |{\vec{a}\cdot \vec{b}|^2=\frac{1}{2}$
• Feb 26th 2013, 04:11 AM
ILikeSerena
Re: Projection Vectors Problem
$\displaystyle \left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 = \frac{|\vec a \cdot \vec b|^2} {|\vec{b}\cdot\vec{b}|^2} ||\vec b||^2 = \frac{|\vec a \cdot \vec b|^2} {||\vec{b}||^4} ||\vec b||^2 = \frac{|\vec a \cdot \vec b|^2} {||\vec{b}||^2}$