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Question:

Find $v_{1}=(a,b,c)$ and $v_{2}=(d,e,f)$ such that $v = v_{1} + v_{2} = (-2,3,-4)$, with $v_{1}$ parallel to $u$ and $v_{2}$ perpendicular to $u$, where $u = (-4,3,-4)$.

Attempt:

Since $v_{1}$ is parallel to $u$ $\therefore $ $v_{1} = ku = (-4k,3k,-4k), k\in \Re $.

Since $v_{2}$ is perpendicular to $u\therefore v_{2}.u=0 \therefore -4d+3e-4f=0$.

Since $v = v_{1} + v_{2} = (-2,3,-4) \therefore$

$$-4k + d=-2\\ 3k + e=3\\ -4k + f=-4\\$$

I don't know where to go from here or how to use $-4d+3e-4f=0$.

Would very much appreciate your help.

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  • $\begingroup$ How about multiply the 1st equation by -4. the second by 3, and and third by -4, and then summing the three equations together? then you'll have the expression that equals to zero, and only k as a single element $\endgroup$ – DanielY Apr 6 '14 at 12:05
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For each $k\in \Re $ that you substitute to the final system of 3 equations you will obtain unique solution

$v_{2}=(d,e,f)$

and

$v_{1} = ku = (-4k,3k,-4k)$

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Try multiplying the 1st equation by -4, the second by 3 and the third by -4. Totally you'll get:

$$16k - 4d = 8$$ $$9k + 3e = 9$$ $$16k - 4f = 16$$

summing all toegther will bring you: $41k - 4d + 3e - 4f = 33$ The expression after $41k$ is $0$, therefore: $41k = 33$

Now you can easily find d, e and f

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  • $\begingroup$ this is only just one special case, but there are infinite number of solutions: one for each k $\endgroup$ – 4pie0 Apr 6 '14 at 12:13
  • $\begingroup$ that's right, he just wanted to know how to use the other equation, so I showed him $\endgroup$ – DanielY Apr 6 '14 at 12:15
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    $\begingroup$ @lizusek that's why I've upvoted your answer, BTW :) $\endgroup$ – DanielY Apr 6 '14 at 12:18
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    $\begingroup$ I've upvoted yours also since indeed this is correct answer to the task described too $\endgroup$ – 4pie0 Apr 6 '14 at 12:21

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