Finding Components, Projections of 3D Vectors

somebodyelse5
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I won't post the actual problem with numbers, I just need some direction. My teacher never went over this part of the webwork in class, and we haven't touched on it in physics either.

"a" and "b" are both 3D vectors.

1.) I am supposed to find the component of "b" along "a"

2.) I am supposed to find the projection of "b" onto "a"

3.) I am supposed to find the projection of "b" orthogonal to "a"


If somewhat could shed some light on what I am actually doing, and maybe give me some direction it would be greatly appreciated.
 
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do you know about dot products? they could be pretty useful here...
 
lanedance said:
do you know about dot products? they could be pretty useful here...

yes, i know how to find dot and cross products. But its not as simple as just finding the dot product is it?
 
Draw a and b with their tails together. Drop a perpendicular from the head of a to the line of vector b forming a right triangle. The component on b of a is the "shadow" of a on b which is the b leg of that triangle and you can see from the picture its length is |b|cos(θ) where θ is the angle between a and b.

Notice that if you make a unit vector out of b, call it bhat that

|a|\cos\theta = |a||\hat b|\cos\theta = a\cdot \hat b

If you multiply that by the unit vector bhat that makes a vector out of the "shadow" and that gives the projection of a on b. Subtracting that projection from a gives the vector forming the other leg of the triangle and that is the orthogonal projection.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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