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Suppose $\Vert\space \vec v\space\Vert$ = 2, $\Vert \space proj_\vec v \space\vec w\space\Vert$ = 5, and the angle between $\vec v$ and $\vec w$ is obtuse. Find $\vec v \cdot\vec w$.

Using what was given, I can deduce that the solution is most likely a negative quantity since the angle between $\vec v$ and $\vec w$ is obtuse.

Also, $proj_\vec v \space\vec w$ = $\vec v \space \cdot \space\vec w\over \Vert v \Vert^2$$\vec v$

Since $\Vert \space \vec v \space \Vert$ = 2, this will result in $proj_\vec v \space\vec w$ = $\vec v \space \cdot \space\vec w\over 4$$\vec v$. At this point, I am unsure what the next step should be to get to the final solution.

Just so you know, this is not a homework problem. I am studying for an exam and would greatly appreciate some clarification on this practice problem.

sYnChris
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  • Hint: $|\vec v \cdot \vec w| = 2 \cdot 5$. Then the sign must be... – dxiv Oct 18 '16 at 01:25
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    One way to geometrically define the dot product is $v \cdot w = \begin{cases}|\operatorname{proj}_vw||v|, & \angle \text{ acute} \ -|\operatorname{proj}_vw||v|, & \angle \text{ obtuse}\end{cases}$ –  Oct 18 '16 at 01:29
  • @dxiv Thanks for the hint. I understand that a negative value for the dot product of two vectors means the angle between them must be obtuse. – sYnChris Oct 18 '16 at 01:38
  • @Bye_World Are you implying that the answer is simply $\vec v \space \cdot \space \vec w$ = $\Vert \space \vec v \space \Vert$ $\cdot$ $\Vert proj_\vec v \space \vec w \Vert$? – sYnChris Oct 18 '16 at 01:38
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    @sYnChris The negative of that, yes. –  Oct 18 '16 at 01:39
  • @Bye_World Ah I see. Thanks for the help! – sYnChris Oct 18 '16 at 01:42

1 Answers1

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Hint: $$\| \operatorname{proj}_v u \| = \left| \frac{u \cdot v}{\| v\|} \right|$$

N. S.
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