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 $u = (-4,3,-4)$.


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.

  • $\begingroup$ 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

For each $k\in \Re $ that you substitute to the final system of 3 equations you will obtain unique solution



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


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ @lizusek that's why I've upvoted your answer, BTW :) $\endgroup$ – DanielY Apr 6 '14 at 12:18
  • 1
    $\begingroup$ I've upvoted yours also since indeed this is correct answer to the task described too $\endgroup$ – 4pie0 Apr 6 '14 at 12:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.