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True/false? $v,w$ are vectors in $\mathbb{R}^2$. The vector $u=\left \langle v,w \right \rangle v -\left \| v \right \|^2w$ lies vertically on $v.$

Can you please explain me how a task like that is solved? It's not homework but a task from an old exercise (you want see it?).

I know that two vectors $a,b$ to be orthogonal we need that $\left \langle a,b \right \rangle=0$

So applying that to this task, we need that $\left \langle u,v \right \rangle=0$, is this correct?

I think this will not be zero if we don't have zero vector, so statement is false? Because it's not satisfied for all vectors I think.

But please I need explanation and how to form things if it's required here at all.

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    $\begingroup$ Have you calculated what $\langle u, v\rangle$ actually is? Do you know how $\|\cdot\|$ relates to $\langle\,\cdot\,,\,\cdot\,\rangle$? $\endgroup$ – Christoph Mar 17 '17 at 18:06
  • $\begingroup$ @Christoph I don't know how to do these calculation with this. You have some good site for me in internet or can show here if it's not too long? $\endgroup$ – tenepolis Mar 17 '17 at 18:07
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    $\begingroup$ You only need to use the fact that $\langle\,\cdot\,,\,\cdot\,\rangle$ is bilinear and symmetric, that means $\langle a, \lambda b+c\rangle = \lambda\langle a,b\rangle + \langle a,c\rangle$ and $\langle a,b\rangle = \langle b,a\rangle$. $\endgroup$ – Christoph Mar 17 '17 at 18:13
  • $\begingroup$ @Christoph Thank you $\endgroup$ – tenepolis Mar 17 '17 at 18:29
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$$\langle v,u\rangle=\langle v,\,\langle v,w\rangle v-\left\|v\right\|^2 w\rangle=\langle v,w\rangle\langle v,v\rangle-\left\|v\right\|^2\langle v,w\rangle=0\implies v\perp u$$

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  • $\begingroup$ Thank you very much. Do you have good page in mind for me? I don't have any source for learn this.. But I learned some from my other questions. If you have good page ready pls share with me. $\endgroup$ – tenepolis Mar 17 '17 at 18:22
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    $\begingroup$ There was two important squares missing, I fixed it. $\endgroup$ – Christoph Mar 17 '17 at 18:41

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