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What angle does the line pass through the points $A[1,1]$ and $B[−3, −3]$ with the vector $v=(−3,\sqrt{3})$?

I proceed as follows:

I determine the vector $AB = u = (-4, -4)$,

then using the formula

$$\cos (x) = \frac{u\cdot v}{\|u\|\cdot\|v\|}$$

I get

$$\cos (x) = \frac{(-4)\cdot (- 3) + \sqrt{3}\cdot (- 4)}{\sqrt{32} \cdot \sqrt{12}}$$

but somehow I can't make adjustments to the result, which is $\frac{5}{12}\pi$.

5xum
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Aaron7
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    Notice I edited your question to improve the formatting. It is strongly advised that you use Mathjax to format your questions on this site - it's like LaTeX for the web. I edited your question this time since you are new, but in future, please format the question yourself. See here for a quick guide: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 5xum Sep 15 '21 at 12:09
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1 Answers1

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Your solution is correct, it just needs some simplification. Your result is

$$\cos(x) = \frac{12-4\sqrt{3}}{\sqrt{32\cdot 12}}$$

while the book result says $x=\frac{5}{12}\pi$. But notice that

$$\cos\left(\frac{5}{12}\pi\right) = \frac{\sqrt{3}-1}{2\sqrt{2}}$$

while $$\frac{12-4\sqrt{3}}{\sqrt{32\cdot 12}} = \frac{4(3-\sqrt{3})}{\sqrt{16}\cdot\sqrt{2\cdot 12}} = \frac{3-\sqrt{3}}{\sqrt{4\cdot 2\cdot 3}} = \frac{\sqrt{3}(\sqrt(3)-1)}{2\cdot\sqrt{2}\cdot\sqrt{3}} = \frac{\sqrt{3}-1}{2\sqrt{2}}$$

5xum
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