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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.

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    @Math Student: I'm sure no one meant to be rude. It's difficult for$a$math beginner to get notation right. Don't worry, if you keep on practicing you'll improve your understanding.2011-02-01

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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?

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    @Arturo: I see, thanks!2011-02-01