I'm trying to find the angle between these two vectors. I know how to find the angle of using the dot product over magnitude of both vectors. However, these two vectors are opposite to each other, creating an angle of 180 degrees and a dot product of -29. Is there a difference between column vectors and row vectors when using this formula?

v = <-5,2> w = <5,-2> (in column form)

What is going on?

  • $\begingroup$ The angle formula is ${\cos \theta} = {{V\cdot W} \over{ ||V || ||W||}}$ which in your case is $-1$. The angle whose cosine is $-1$ is $180^\circ$. Sometimes it is written as $AB^T$ if $A,B$ are rows or as $A^TB$ if they are column vectors, or ... $\endgroup$ – Maesumi Feb 16 '13 at 0:47
  • $\begingroup$ No difference at all. $\endgroup$ – Ross Millikan Feb 16 '13 at 0:47
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    $\begingroup$ Public service announcement: please do not get in the habit of numerically computing the angle using $\arccos$. For 2D and 3D vectors use $\operatorname{atan2}( |u\times v|, u\cdot v).$ $\endgroup$ – user7530 Feb 16 '13 at 0:51
  • $\begingroup$ @user7530 -- why is that? $\endgroup$ – bubba Feb 16 '13 at 1:06
  • $\begingroup$ I had a feeling that there was no difference. I got stumped when I found out that the answer had to be in radians and not in degrees which led to my belief that I was in the wrong... $\endgroup$ – Dimitri Feb 16 '13 at 1:07

\begin{align} u&=(-5,2)^T\\v&=(5,-2)^T\\ \cos(\theta)&=\dfrac{u^Tv}{||u||\times ||v||}\\ &=\dfrac{-29}{\sqrt{29}\times \sqrt{29}}\\ &=-1\\ \implies \theta&=180 \end{align}

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  • $\begingroup$ Hi, this is the answer I got, however it is wrong in my WebWork. I don't understand why though. $\endgroup$ – Dimitri Feb 16 '13 at 0:48
  • $\begingroup$ @DimitriTopaloglou What is "WebWork"? $\endgroup$ – Inquest Feb 16 '13 at 0:48
  • $\begingroup$ WebWork is a common place where students across Universities do their assignments online. $\endgroup$ – Dimitri Feb 16 '13 at 0:49
  • $\begingroup$ Maybe they want the answer in radians? $\endgroup$ – user7530 Feb 16 '13 at 0:51
  • $\begingroup$ Perhaps the answer is in radians... I'll give that a try. $\endgroup$ – Dimitri Feb 16 '13 at 0:51

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