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I have such a question from the exercises given by the professor

$\vec{u}$ and $\vec{v}$ are two vectors on the plane and $\vec{w}=\vec{u}+\vec{v}$. We suppose that $||\vec{u}|| = 3$ and $||\vec{v}|| = 5$ and $\vec{w} \vec{u} = 0$. Please calculate $\vec{u} \vec{v}$

According to the answer key, the procedure is as following

  • $\vec{u} \vec{v}$ = $\vec{u}(\vec{w}-\vec{u})$
  • =$\vec{u}\vec{w}-\vec{u}\vec{u}$
  • =$0-||\vec{u}||^2$
  • =-9

However, from my understand, the formula of dot product should be $\vec{u} \vec{v} = ||\vec{u}||||\vec{v}||cos\theta$

So why the $cos\theta$ is not calculated here?

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  • $\begingroup$ Do you know how to compute dot products using the coordinates of the vectors? $\endgroup$ – Carl Christian Oct 31 '19 at 22:07
  • $\begingroup$ Yes, but the problem is that the coordinates are not given. $\endgroup$ – Yan Zhuang Oct 31 '19 at 22:09
  • $\begingroup$ The dot product of vectors with coordinates is easy as $u_1v_1+u_2v_2$. However, that would require me actually knowing the coordinates of each vector no? $\endgroup$ – Yan Zhuang Oct 31 '19 at 22:23
  • $\begingroup$ You don't immediately know the angle between $u$ and $v$, so you can't use that formula to find the dot product. But now that you have the dot product, you could find that angle if you wanted. $\endgroup$ – Ned Oct 31 '19 at 22:29
  • $\begingroup$ @Ned I am sorry, but I did not understand what you mean? I am not trying to find the angle. The answer provided did not calculate the angle (While the dot product of geometric vector requires the calculation of angle). So I am not sure why, in this question, they did not calculate and get the answer for the dot product $\endgroup$ – Yan Zhuang Oct 31 '19 at 22:40
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You are given $\vec w \cdot \vec u =0$.

Note $\vec w \cdot \vec u = (\vec u + \vec v) \cdot \vec u$

$= \vec u \cdot \vec u + \vec v \cdot \vec u $

(distributivity of the dot product)

$=\vec u \cdot \vec u + \vec u \cdot \vec v $

(commutativity of the dot product)

$= | \vec u|^2+\vec u \cdot \vec v$

So, $| \vec u|^2+\vec u \cdot \vec v=0$

$3^2+\vec u \cdot \vec v =0$

$\therefore \vec u \cdot \vec v = -9$

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  • $\begingroup$ That is to say, the answer to this question is wrong? Because there are actually three others sub-question to this in which all of them involve the result from this first question, and all of them uses -9 $\endgroup$ – Yan Zhuang Oct 31 '19 at 22:47
  • $\begingroup$ It would appear your answer key is wrong as long as you presented the question exactly right. Check. $\endgroup$ – Deepak Oct 31 '19 at 22:49
  • $\begingroup$ Okay. I will write to the professor to verify! Thank you very much! $\endgroup$ – Yan Zhuang Oct 31 '19 at 22:51
  • $\begingroup$ Yes, you should. But first ensure your question is written correctly here as I said. Because your question stipulates wv=0, but your answer key assumes uw=0. $\endgroup$ – Deepak Oct 31 '19 at 22:55
  • $\begingroup$ Or the answer key swapped the norms of the two vectors. $\endgroup$ – amd Oct 31 '19 at 22:56

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