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Im having problem determining cosines. Does anyone know how to solve it? Can anyone provide some hint or full solution? Please Help !

Data are the vectors $x$ and $y$, where $\| x \| = 3$, $\|y\| = 4$, and $xy = 2$. Determine $\| x - 2y \|$, $\| x + 3y \|$ and $cos (∡ (x - 2y, 3y + x))$.

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    What have *you* been able to do with it so far? – 2017-02-08
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    Does "$\;xy=2\;$" mean the scalar (dot) product of the vectors $\;x,y\;$ equals two? – 2017-02-08
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    ||x-2y|| = -5 ||x+3y||=9 cos(x+y,x+3y) = 2/((-5) * 9) <-- but teacher said its wrong way – 2017-02-08
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    xy => (x|y) = ||x|| * ||y|| cos (x,y) – 2017-02-08
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    I didn't know how to interpret the symbol $xy>=2$. – 2017-02-08
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    it pasted adittional sign its xy ==> (x|y) – 2017-02-08
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    @DaveM Perhaps you could learn something important in the next link: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-02-08
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    This will also be useful, thanks :] – 2017-02-08

1 Answers 1

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For example, and denoting by $\;\langle x,y\rangle\;$ the dot product of $\;x,y\;$:

$$\left\|x-2y\right\|^2=\langle x-2y,\,x-2y\rangle=\langle x,x\rangle-4\langle x,y\rangle+4\langle y,y\rangle=9-4\cdot2+4\cdot16=65$$

Do the rest, and also remember that if $\;\theta=\angle(a,b)\;$ ,then

$$\cos\theta=\frac{\langle a,b\rangle}{\left\|a\right\|\left\|b\right\|}$$

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    You need $-4\langle x,y \rangle$ – 2017-02-08
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    @user160738 Good catch, thanks. – 2017-02-08
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    Thank you i will try with this :) – 2017-02-08
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    @DonAntonio But shouldn't it be that 4⟨y,y⟩ = 4 * ||4||^2 so its 65 not 17? – 2017-02-08
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    @DaveM True. Good catch! – 2017-02-08
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    @DonAntonio so is it correct that ||x+3y|| =171 ? and cos(∡(x−2y,3y+x)) = 0.000179937 ? – 2017-02-08
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    my mistake ||x+3y|| = 189 right? and then cos(∡(x−2y,3y+x)) = 0.0001628002? – 2017-02-08