Linear Algebra - Find the vectors

cris623
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Homework Statement



For u=(−5 −1 −13) and v=(2 1 4), find the vectors u1 and u2 such that:
(i) u1 is parallel to v
(ii) u2 is orthogonal to v
(iii) u = u1 + u2

The Attempt at a Solution



Since u2 is perpendicular to v I used the dot product set equal to zero:

(x y z) . ( 2 1 4) = 2x + y + 4z=0

Past that I'm not too sure.
I can see that u1 and u2 are also perpendicular to each other and u1 will be some t(2 1 4) but other than that, I have no idea

Thanks in advance
 
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Your attempt at finding u2 actually involved using the dot product. A cross product would have produced a vector.
 
Yeah you're right. I meant to put cross product.
 
I would NOT use a cross product because you don't have two vectors you want u2 orthogonal to.

Any vector parallel to (-5, -1, -13) is of the form (-5a, -a, -13a) for some number a.
any vector, (x, y, z) orthogonal to (2, 1, 4) satisfies 2x+ y+ 4z= 0.

u= u1+ v means that (-5a, -a, -13a)+ (x, y, z)= (-5, -1, -13) so that (-5a+ x, -a+ y, -13a+ z)= (-5, -1, -13) which means you want to solve -5a+ x= -5, -a+ y= -1, -13a+ z= -13. with 2x+ y+ 4z= 0. From that last equation, y= -2x- 4z. Putting that in for y in those equations gives three equations to solve for x, y, and a.

(That turns out to have a rather trivial solution.)
 
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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