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:
$\displaystyle \text{Proj}_{ab} = \mathbf a \cdot \hat{\mathbf b}$
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?
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.
$\displaystyle \left\|\frac {\vec{a}\cdot \vec{b}}{\vec{b}\cdot\vec{b}}\;\vec{b} \right\|^2 = \frac{|\vec a \cdot \vec b|^2} {|\vec{b}\cdot\vec{b}|^2} ||\vec b||^2 = \frac{|\vec a \cdot \vec b|^2} {||\vec{b}||^4} ||\vec b||^2 = \frac{|\vec a \cdot \vec b|^2} {||\vec{b}||^2}$