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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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3 Answers

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

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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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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