I don't understand who projects on who. How is the following a projection on anything? I don't see how a vector of lenght 6 can possibly result of a projection here. If I project C down to the x-axis, it lands on "2". If I project B upwards towards C it will result in something along "3". Why would it be "6" ?!
How is the dotproduct a projection?
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0Why should you "see how a vector of length 6 can possibly result of a projection here"? – 2012-11-07
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0Zhang, the result of the dotproduct is 6. Everywhere I have read, people say that the dotproduct is a projection of one vector to another. I just don't see how that's possible when the result is 6 here. – 2012-11-07
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0The idea of projection works only when the vector being projected onto has unit length. – 2012-11-07
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0@Zhang [Helpful?](http://en.wikipedia.org/wiki/Dot_product#Geometric_interpretation) – 2012-11-07
1 Answers
The dot product $a\cdot b$ tells you something about the length of the projection of $a$ onto $b$, (or $b$ onto $a$, since the relationship is symmetric.)
That is, if you decompose $a=a_\perp+a_\parallel$ where $a_\perp$ is perpendicular to $b$ and $a_\parallel$ is parallel to $b$, then $a\cdot b=|a_\parallel|$.
Edit: Actually it will not be exactly that. A corrected version of what I was thinking of follows. Again, decompose $a=a_\perp+a_\parallel$ where $a_\perp$ is perpendicular to $b$ and $a_\parallel$ is parallel to $b$. If $b$ has unit length, then $a\cdot b$ will give the length of $a_\parallel$. If $b$ doesn't have unit length, then the dot product reports $|b||a_\parallel|$ or $-|b||a_\parallel|$, depending on if the angle between $a$ and $b$ is acute or obtuse, respectively.
I would also have to modify this explanation to account for situations where there the dot product is negative by referring to a "signed length". At any rate, I think this is the way to link projection and the dot product.
So, for example, if $a$ and $b$ are perpendicular, then the projection of $a$ onto $b$ is a single point (i.e. a length zero vector) and so $a\cdot b=0$ in that case.
In your example, $a\cdot b=6$, which is, depending on how you look at it, the length of $b$ times the length of $a_\parallel$, ($3\cdot 2$) or the length of $a$ times the length of $b_\parallel$ ($2\sqrt 2 \cdot \frac{3}{\sqrt{2}}$).
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0How did you come up with the 3/sqrt(2)? Your explanation helped a lot, but I was still put off by the word "projection". ...change vector 'a' to (0,5/0,5) for example. Now the resulting dotproduct would be 1.5, which is half of b's length. If I think of it as two textures being multiplied together, being projected onto each other like with a film projection... maybe that makes more sense? It seems a little crazy, but okay. At least with textures, it makes more sense that if you multiply another texture's red channel by 0, it will also be 0, so you basically "multiply" the textures. – 2012-11-07
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0@Blub "project" $b$ onto $a$ by dropping a perpendicular back toward $a$. Thus $b$ forms the hypoteneuse of a 45-45-90 triangle, and the side parallel to $a$ must have length $3/\sqrt{2}$. – 2012-11-07
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0Projecting $a$ onto $b$ is much simpler: the perpendicular drops vertically straight down onto $b$. – 2012-11-07