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With the vectors $u,v$ and $w$ being vectors in $\mathbb{R}^3$ with $u$ and $v$ given, how do I approach the problem of calculating $w$ given the following: $$\langle2u,3w-5v\rangle = -4$$ Due to linearity, this can be transformed to $$2 ( 3 \langle u,w\rangle- 5\langle u,v\rangle) = -4$$

What next?

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That is not enough information to find $w$ uniquely. Just plug-in $w=(a,b,c)$, calculate the inner product - and you will have one equation with three variables - $a,b,c$ –  Dennis Gulko Apr 25 '13 at 10:28
    
In fact, this is the equation of a straight line. –  Raskolnikov Apr 25 '13 at 10:29

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Yes, so you get $$6\,\langle u,w\rangle = 10\,\langle u,v\rangle\,-4\,, \\ \langle u,w\rangle = \frac{5\,\langle u,v\rangle\,-2}3\,,$$ where the right hand side is a given constant. If it was $0$, then the solution set would be $$\{ w\in\Bbb R^3\,\mid\, u\perp w\}=:\,u^\perp\,,$$ i.e. the orthogonal plane to $u$. (Or, the whole space in case $u=0$.) If the constant is not $0$, it just means that the solution set is parallel to the plane $u^\perp$, and arbitrarily finding one vector from it (one particular solution) will determine the whole parallel plane.

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thanks so far. Building on your post, given the right side != 0 , solving up the left side yields me an equation with three unkowns (w_1 , w_2, w_3) and this equation is equal to the solution of the right side. How can I even solve that? –  Rickyfox Apr 25 '13 at 11:11
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What is that equation? I think, you don't need to 'solve' that any further, because that equation gives all the possible $w$'s. Or, as I suggested, just pick any concrete solution $w_0$ and then you can state that the solution set is $w_0+u^\perp$. –  Berci Apr 26 '13 at 10:13

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