1
$\begingroup$

Find u, v and w such that all are unit vectors and $\textbf{u} + 2\textbf{v} + \textbf{w} = 0$.

I can guess a solution, but u and w are the same in my solution, of course the problem has not made it explicit that u and w must be different, but I'm not sure if my solution is a good one: u and w = $(cost, sint, 0)$, and v= $(-cost, -sint, 0)$. What are other possible answers? Also another issue is that I've guessed these values for u, w and v, is there a way to solve such an equation?

$\endgroup$
5
  • $\begingroup$ I think your solution is a good one $\endgroup$ Jan 29, 2021 at 19:28
  • $\begingroup$ I knew I would get this comment :D But my main question is, is guessing the way to solve this equation? $\endgroup$
    – aderchox
    Jan 29, 2021 at 19:29
  • 1
    $\begingroup$ You're never going to be able to solve in the sense you're thinking, since there are infinitely many solutions. But here's a start: What do you know about $\|\mathbf u + \mathbf w\|$? Can you work that out? $\endgroup$ Jan 29, 2021 at 19:29
  • $\begingroup$ @TedShifrin That would help? u+w=-2v, so ||u+w|| would be ||-2v|| = 2. $\endgroup$
    – aderchox
    Jan 29, 2021 at 19:31
  • $\begingroup$ Good. Keep going. I wrote a bit more as a hint in the answer. $\endgroup$ Jan 29, 2021 at 19:33

2 Answers 2

3
$\begingroup$

HINT: If $\mathbf u+\mathbf w$ is twice a unit vector, you should be able to deduce that $\mathbf u\cdot\mathbf w = 1$. Now what does that tell you?

$\endgroup$
5
  • $\begingroup$ Yes I could deduce that. This means that they are in the same direction, also both are unit vectors, so they must be equal. Thanks. $\endgroup$
    – aderchox
    Jan 29, 2021 at 19:43
  • $\begingroup$ Of course if I may ask a small question too, I had to write a small proof for what you said, but how did you deduce it so fast? From experience, or there's an intuitional relation for ||u+w||=2 => u.w=1 as well? $\endgroup$
    – aderchox
    Jan 29, 2021 at 19:47
  • 1
    $\begingroup$ Oh, no, it requires proof. Expand $\|\mathbf u+\mathbf w\|^2 = (\mathbf u+\mathbf w)\cdot(\mathbf u+\mathbf w)$. (Alternatively, you have to use the law of cosines, but I prefer the dot product.) $\endgroup$ Jan 29, 2021 at 19:49
  • $\begingroup$ Great, thanks. I proved it another way, by writing the components and doing the dot product. Thanks. $\endgroup$
    – aderchox
    Jan 29, 2021 at 19:51
  • 1
    $\begingroup$ Oh, good, but try to avoid writing things out in terms of components unless absolutely necessary. Try to use the power of the linear algebra. :) $\endgroup$ Jan 29, 2021 at 19:56
2
$\begingroup$

I think it is not possible to find another solution (where $u\ne w$).

Assume all vectors are (pairwise) different. Since $u+2v+w=0$, they form a triangle with sides $|u|=1$, $|2v|=2$ and $|w|=1$. This is not possible unless $u=w=-v$ (in a nondegenerate triangle, the sum of any two sides must be strixctly greater than the other).

$\endgroup$
9
  • 1
    $\begingroup$ Did you mean $u=w=-v$ (rather than $-2v$)? $\endgroup$ Jan 29, 2021 at 19:42
  • $\begingroup$ Yes, thanks @J.W.Tanner $\endgroup$ Jan 29, 2021 at 19:44
  • $\begingroup$ I edited my comment in response to your edit. And what about $u=w=(0,0,1)$? $\endgroup$ Jan 29, 2021 at 19:44
  • $\begingroup$ Then $v=(0,0,-1)$ $\endgroup$ Jan 29, 2021 at 19:48
  • 1
    $\begingroup$ @J.W.Tanner I guess Tito meant it's not possible to find any solution other than keeping u=w. (Excuse me for interpreting your answer before yourself Tito) $\endgroup$
    – aderchox
    Jan 29, 2021 at 19:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .