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Let u and v be unit vectors. Prove that u+v is perpendicular to u − v.

What I did is:

Since they are unit vectors: $|u|$=$|v|$=1

If they are perpendicular, then $(u+v)$.$(u-v)$=$0$

Ans: $|u+v|$.$|u-v|$.$\cos\theta$

= (1+1).(1-1).$\cos\theta$

=$0$

proved.

Is this right?

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    $\begingroup$ $$|u-v| \neq |u| - |v|$$ $\endgroup$ Commented Dec 1, 2014 at 14:38
  • $\begingroup$ No, because |u - v| is not |u| - |v| which you have assumed here. You need to note that $u.u = |u|^2$ to get it right. $\endgroup$
    – Paul
    Commented Dec 1, 2014 at 14:39
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – user137731
    Commented Dec 1, 2014 at 14:39

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No, it is not right, because $|u+v|$ is not $|1+1|$.

Try this, instead: $$(u+v)\cdot(u-v)=u\cdot u+u\cdot v-v\cdot u-v\cdot v=\ldots$$

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  • $\begingroup$ Thank you for answering. I'm new to vectors and they seem difficult to me. Could you please complete your answer? $\endgroup$
    – Mohamed
    Commented Dec 1, 2014 at 14:49
  • $\begingroup$ For a unit vector $u$, product $u\cdot u$ is $1$. The same is true for $v\cdot v$. So, it remains to be shown that $u\cdot v - v\cdot u$ is in fact $0$. $\endgroup$ Commented Dec 1, 2014 at 14:55
  • $\begingroup$ @MohamedAl-Qabtan Just remember that $|u|^2 = u \cdot u$. $\endgroup$
    – user137731
    Commented Dec 1, 2014 at 14:59

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