Linear Algebra - Find the vectors

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Homework Help Overview

The problem involves finding two vectors, u1 and u2, based on given vectors u and v in a linear algebra context. Specifically, u1 must be parallel to v, while u2 must be orthogonal to v, and their sum must equal u.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use the dot product to establish the orthogonality condition for u2, while expressing uncertainty about the next steps. Some participants suggest reconsidering the use of the cross product, while others clarify that a cross product is not applicable in this context.

Discussion Status

Participants are exploring different methods to approach the problem, with some providing guidance on how to express the vectors and set up the equations. There is an ongoing discussion about the relationships between the vectors and the equations that need to be solved.

Contextual Notes

There is a mention of the need to solve a system of equations derived from the relationships between u1, u2, and the original vectors, indicating that the problem may have multiple interpretations or approaches.

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.)
 

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