I want to find the vector that meets the following:
$$X\parallel (2,1,-1)$$
$$X*(2,1,-1)=3$$
what I did so far is :
$$2x+y+z=3$$
I know that parallel vectors the angle is $180$ or $0$.
how to continue from here?
Note: X is vector.
Thanks!
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Ofir Attia
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What $X * (2,1,-1)$ does mean? – dtldarek Jul 19 '13 at 09:20
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@dtldarek: presumably, inner product – SBF Jul 19 '13 at 09:24
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I'd suggest taking another look at that $2x+y+z=3$ thing. Then I'd interpret parallelism as "is a constant multiple of". – Gerry Myerson Jul 19 '13 at 09:26
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The easiest way is to notice that $X\parallel Y$ iff $X = a Y$ for some $a\in \Bbb R$. That is, you know that $$ X = (2a,a,-a) $$ and that $2a\cdot 2 + a\cdot 1+(-a)\cdot (-1) = 3$ which will give you $a$.
An "alternative" way which is not really an alternative is so mention that you have two conditions: $X\parallel Y$ and $X\cdot Y = 3$. The first is equivalent to $X = aY$ as we already discussed, so the second now turns to be $$ 3 = X\cdot Y = aY\cdot Y = a\|Y\|^2 \iff a = \frac3{\|Y\|^2} $$ where $\|Y\|$ is the length of $Y$, which you can find using squares and square roots.
SBF
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Ok, so for example I have$$ X\parallel (2,1,-1)$$ and the second thing is $$|X|=3 $$ is mean that is going to be $x^2+y^2+z^2=9$ than I insert the X as you did? – Ofir Attia Jul 19 '13 at 09:36
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@GerryMyerson: indeed, but I'm pretty sure that OP has more than this single problem on the same topic, so I saw no harm of showing how would it work till the end. – SBF Jul 19 '13 at 09:53
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@OfirAttia: indeed. Note, however, that in problem you came up with it holds that $|aY| = |-aY|$, so that you'll have two solutions. – SBF Jul 19 '13 at 09:54
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@Ilya I think you want the denominator of your final fraction to be $|Y|^2$ and not $|Y^2|$. – Daryl Jul 19 '13 at 09:59
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