# An equation of unit vectors

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?

• I think your solution is a good one Jan 29, 2021 at 19:28
• I knew I would get this comment :D But my main question is, is guessing the way to solve this equation? Jan 29, 2021 at 19:29
• 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? Jan 29, 2021 at 19:29
• @TedShifrin That would help? u+w=-2v, so ||u+w|| would be ||-2v|| = 2. Jan 29, 2021 at 19:31
• Good. Keep going. I wrote a bit more as a hint in the answer. Jan 29, 2021 at 19:33

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?

• 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. Jan 29, 2021 at 19:43
• 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? Jan 29, 2021 at 19:47
• 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.) Jan 29, 2021 at 19:49
• Great, thanks. I proved it another way, by writing the components and doing the dot product. Thanks. Jan 29, 2021 at 19:51
• 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. :) Jan 29, 2021 at 19:56

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).

• Did you mean $u=w=-v$ (rather than $-2v$)? Jan 29, 2021 at 19:42
• Yes, thanks @J.W.Tanner Jan 29, 2021 at 19:44
• I edited my comment in response to your edit. And what about $u=w=(0,0,1)$? Jan 29, 2021 at 19:44
• Then $v=(0,0,-1)$ Jan 29, 2021 at 19:48
• @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) Jan 29, 2021 at 19:53