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Find the angle α between the vectors $$\begin{bmatrix}3 \\ 5 \end{bmatrix}and \begin{bmatrix}-2 \\ 3 \end{bmatrix} $$

I found 64.654, but apparently is wrong from what the webwork says. Can anyone check if they get the same answer ? or the calculation i should be using.

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That's what I get too (in degrees). You may want to answer in radians. –  mrf Feb 21 '13 at 22:49
    
@mrf, I converted to radian with google, I get 0.0174532925 which is wrong. Also tried setting my calculator to radians and the number 1.13 which is also wrong Y_Y becoming desperate –  Reza M. Feb 21 '13 at 22:54
    
Two professional mathematicians already told you that you're right, Reza, so odds are you are right and the book is wrong. Believe me, this happens way more times than beginner students seem to be ready to accept. –  DonAntonio Feb 21 '13 at 23:04

3 Answers 3

up vote 6 down vote accepted

Assuming the vectors are "anchored" at the origin and that we have the usual inner product in $\,\Bbb R^2\,$ , we get that the wanted angle $\,\theta\,$ is given by

$$\cos\theta=\frac{\binom{3}{5}\cdot\binom{-2}{3}}{\left|\left|\binom{3}{5}\right|\right|\;\left|\left|\binom{-2}{3}\right|\right|}=\frac{-6+15}{\sqrt{34}\sqrt{13}}=\frac{9}{\sqrt{442}}\Longrightarrow \theta=\arccos\frac{9}{\sqrt{442}}=64.65^\circ$$

and thus your answer is correct...unless the assumptions in the first part are different, of course.

Another possibility is they want the angle in radians, so after the known conversion it'd be $\,\theta=1.13\,$ radians

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I tried the radian value as well, and both are wrong T_T –  Reza M. Feb 21 '13 at 22:55
    
Well Reza, another very possible option is that the webwork is wrong...;) –  DonAntonio Feb 21 '13 at 23:02
    
alright, ty, won't waste further time on it. –  Reza M. Feb 21 '13 at 23:13

$$\vec{u} \cdot \vec{v} = \|u\| \|v\| \cos \theta$$

The dot product is 9, and magnitudes are $\sqrt{34}$ and $\sqrt{13}$, so

$$\cos \theta = \frac{9}{\sqrt{34 \cdot 13}}$$

so $\theta = \arccos \frac{9}{\sqrt{34 \cdot 13}} = 64.65$ degrees...

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Can't seem to get it working. –  Reza M. Feb 21 '13 at 22:55

Hello: you can use the dot product between vectors

$\theta = \arccos(\frac{3\cdot(-2)+5\cdot(3)}{\sqrt{(3^2+5^2)(2^2+3^3)}})$

if you use wolfram alpha for to get the result

Ans: $\theta = 1.2898983977450732031495202240229012130410696446569681$

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The value still wrong. I'm just going to speak to the teacher. Thanks. –  Reza M. Feb 21 '13 at 23:15
    
This computation is incorrect –  mrf Feb 21 '13 at 23:17
    
It's not $3^3$ but $3^2$ in the denominator's right factor under the square root.... –  gt6989b Feb 21 '13 at 23:46

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