What are scalar multiples and projections in vector operations?

lypaza
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[PLAIN]http://img62.imageshack.us/img62/5319/49966749.png

What is the scalar multiples of a vector actually?
I was thinking L = c[2 1 2]T
Then I looked for projection of v on L. But I got c in my answers which are not supposed to be...
 
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lypaza said:
[PLAIN]http://img62.imageshack.us/img62/5319/49966749.png

What is the scalar multiples of a vector actually?
I was thinking L = c[2 1 2]T
Then I looked for projection of v on L. But I got c in my answers which are not supposed to be...
Every vector in L is some scalar multiple of <2, 1, 2>. The line goes through the origin - the zero multiple of this vector is 0<2, 1, 2> = <0, 0, 0>, a vector that starts and ends at the origin. The line goes through the point (-4, -2, -4), which you can get by taking the -2 multiple of the vector.

For the reflection of v in the line, you want to find another vector w that is in the same plane as v and L, but is on the opposite side of L, and makes the same angle with L.
 
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So I have to find angle theta between v and L, and then find vector w with negative theta?
I also have to find the plane of v and L by cross product of v and L ...
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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