What are scalar multiples and projections in vector operations?

lypaza
Messages
17
Reaction score
0
[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...
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
Last edited by a moderator:
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 ...
 
Last edited:
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...

Similar threads

Replies
1
Views
1K
Replies
4
Views
3K
Replies
12
Views
2K
Replies
48
Views
4K
Replies
5
Views
1K
Back
Top