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Find a unit vector that is orthogonal to both $u = (1, 0, 1)$ and $v = (0, 1, 1)$.

I am lost on how to find one orthogonal to both.. I used the unit vector rule:

vector $= (1/||v||)v$ on both vectors to get $(-1/\sqrt2,0,-1/\sqrt{2})$ and $(0,-1/\sqrt2,-1/\sqrt2)$ after making them negative to make them oppositely directed.. but I can't figure out how to find one that is orthogonal to both..

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    $\begingroup$ Take the cross product. $\endgroup$
    – Gaussler
    Jun 26, 2014 at 5:40
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    $\begingroup$ Alternate hint to the cross product: Choose a vector $(a,b,c)=w$ such that $w\cdot u=0$ and $w\cdot v=0$. $\endgroup$
    – user18862
    Jun 26, 2014 at 5:41
  • $\begingroup$ Did you do the Gram-Schmidt process yet? $\endgroup$
    – GEdgar
    Feb 21, 2017 at 2:58

3 Answers 3

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$u$ and $v$ are orthogonal if $u\cdot v=0$.

you want a vector $(a,b,c)$ such that $(a,b,c)\cdot (1,0,1)=0$ and $(a,b,c)\cdot (0,1,1)=0$. That is $a+c=0$ and $b+c=0$. There are many possible solutions for $a,b,c$ which satisfy both of these equations. For example, $a=1,b=1,c=-1$ works.

Now multiply $(1,1,-1)$ by $1/$(its norm) to get the unit vector $(1/\sqrt 3,1/\sqrt 3,-1/\sqrt 3)$.

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  • $\begingroup$ @JL glad I could help $\endgroup$ Jun 26, 2014 at 5:48
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one can use the cross product to find the vector orthogonal to two vectors. Divide by the magnitude to find the unit vector same result. The method intially used is an around about method of achieving it.

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  • $\begingroup$ So, were you going to answer the question? $\endgroup$
    – The Count
    Feb 21, 2017 at 3:16
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Hint: What can you say about the cross product: $u \times v$?

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  • $\begingroup$ havent got that far yet $\endgroup$
    – J L
    Jun 26, 2014 at 5:44

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