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Find 2 vectors u and v such that they are perpendicular one of the vector is twice the magnitude of the other. And the sum vector of u and v is [6,8]
I did:
Let u=[a,b]
Let v=[c,d]
Let |u|=2|v
u.v=ac+bd=0
|u+v|=|u|^2 + |v|^2
But |u|=2|v|
|u+v|=5|v|^2
5|v|^2=100
|v|^2=20
Im stuck after this
You forgot to write ^2 on |u+v|
Anyways, can you write |v|^2 in terms of a, b, c, and d? Do you know what |u|^2 is? Have you used the fact that u + v = [6, 8] yet?
So
|u|^2+|v|^2=|u+v|^2
But since |u|^2=4|v|^2
5|v|^2=36+64
|v|^2=20
now where can I go?
Can you write |v|^2 in terms of a, b, c, and d? Do you know what |u|^2 is? Have you used the fact that u + v = [6, 8] yet?
|v|^2 = c^2+d^2
|v|^2 = 1/2 (a^2 + b^2)
1/2(a^2+ b^2)= c^2 + d^2
Is that what you mean?
Can you think of anything better you can do with those first two equatinos?
Since |v|^2=2|u|^2,
|v|^2 = c^2+d^2
|u|^2 = 1/2 (c^2 + d^2)
|u|^2 + |v|^2 = 100
3/2 (c^2 + d^2) = 100
I dont know where Im headed
what formulas do you have involving |v|^2?
Projection of u on v
u on v = [u.v/|v|^2] |v|
I guess I should have asked this first off...
You know that your goal is to find 4 equations involving only a, b, c, d which you can solve, right?
4 equations?? I didnt know that...I've been focusing on finding a and b
Well, you have 4 unknowns; a, b, c, and d. In general, you need 4 equations to solve for all 4 of them.
Sometimes you can get lucky and you can find two equations that involve only a and b, but in general that won't happen (and I'm pretty sure it doesn't here)...
You've already found one good equation:
ac + bd = 0
You just need 3 more! You can get 2 more equations out of what you've told me about |v|^2...
Oh, BTW, if |u| = 2|v|, then |u|^2 = 4|v|^2
Ill try this and post a little later, thanks man
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