How do I find Orthogonal Projection given two Vectors? Given two vectors, a and b, how do I find the Orthogonal Projection?
I've already found the Scalar and Vector Projections.
\begin{align*}
\text{Scalar}&:\quad  \frac{-90 + -25 + 24}{\sqrt{9^2+5^2+8^2}};\\
\text{Vector}&:\quad \left(\left(\frac{-91}{\sqrt{170}}\right)\left(\frac{-9}{\sqrt{170}}\right), \left(\frac{-91}{\sqrt{170}}\right)\left(\frac{-5}{\sqrt{170}}\right), \left(\frac{-91}{\sqrt{170}}\right)\left(\frac{8}{\sqrt{170}}\right)\right).
\end{align*}
Here's the original question: Let a = (-4, -8, -4) and b = (-3, -2, 0) be vectors. Find the scalar, vector, and orthogonal projections of b onto a.
Thank you for your time and help.
 A: About the vector projection of $\vec{b}$ onto $\vec{a}$. I am taking its definition from here http://en.wikipedia.org/wiki/Vector_projection where it is given as $(\vec{b} \cdot \vec{e}_a ) \cdot \vec{e}_a$ where $\vec{e}_a$ is the unit vector in $a$-direction.
First for some explanation:
What does the scalar product "$\cdot$" do? If you look at this picture: http://upload.wikimedia.org/wikipedia/commons/3/3e/Dot_Product.svg you can see that it gives you the length of $A$ in $B$ direction. What does it mean? You can picture it like this:
If the sun shines onto the vectors straight from above, the shadow of $A$ cast onto $B$ is exactly the length of $A$ in the direction of $B$.
The scalar product is defined to be $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \Theta$ so you know how to calculate  this length: $|A| cos \Theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$. In your case $\vec{B} = \vec{e}_a$ is a unit vector so its length is one and therefore you get $\vec{b} \cdot \vec{e}_a = |\vec{b}| \cos \Theta$ which is the length of $\vec{b}$ in the direction of $\vec{a}$.
A vector always consists of a length and a direction so now you need to add a direction to the length you just computed. But the direction is the direction of $\vec{a}$. How do you get the direction only without its length? You make it a unit vector like this: $\vec{e}_a = \frac{1}{|\vec{a}|} \vec{a}$.
You see that $length$ $\cdot$ $direction$ is $(\vec{b} \cdot \vec{e}_a ) \cdot \vec{e}_a$ in this case.
Hope this helps. As for the other two cases, scalar projection and orthogonal projection, I don't know what they are. Maybe you could post their definitions given in your lecture here?
