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I am in calculus 3 and the chapter on vectors (cross product) one of the properties is:
u x v = 0 if and only if u + v are scalar multiples of each other
I need to prove u x v = 0
A hint we have is:
u x v = 0
assume -> deriv ; and then take it back the other way.
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are u and v both 3 dimensional?
X'(y-Xβ)=0
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As I believe George was hinting to, write out the vectors piecewise, do all the necessary operations, and the answer should appear.
For example, going the one way would be:
let u = <x, y, z>
v = <ax, ay, az>
Now compute the cross product.
The other way, start with two vectors:
u = <x, y, z>
v = <a, b, c>
And compute the cross product again, and set it to 0. Show that this resulting vectors shows that you have write u as a scalar multiple of v.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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alternatively, rather than doing it piecewise, you could just write:
this will be 0 when sinθ = 0, which is when the two vectors a and b are collinear. if they are collinear then they can only be multiples of eachover. The only other time the two vectors can give a 0 vector, is if one or both of them is a 0 vector, in which case they need not be collinear, but then the other is still a multiple of the other by 0.
Last edited by luca-deltodesco (2007-06-27 18:48:18)
The Beginning Of All Things To End.
The End Of All Things To Come.
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