# Given that |vec(n)|=1,|vec(m) |=sqrt(3) And that the angle between the vectors |vec(m)| and |vec(n)| is 30^(circ). We will define vec(a) =vec(m)−vec(n) , vec(b) =vec(m) +vec(n) calculate the area of ​​the triangle created by vectors vec(a) ,vec(b).

Given that $|\stackrel{\to }{n}|=1,|\stackrel{\to }{m}|=\sqrt{3}$ And that the angle between the vectors $|\stackrel{\to }{m}|$ and $|\stackrel{\to }{n}|$ is ${30}^{\circ }$
We will define $\stackrel{\to }{a}\phantom{\rule{mediummathspace}{0ex}}=\stackrel{\to }{m}\phantom{\rule{mediummathspace}{0ex}}-\stackrel{\to }{n}\phantom{\rule{mediummathspace}{0ex}},\phantom{\rule{mediummathspace}{0ex}}\stackrel{\to }{b}=\stackrel{\to }{m}+\stackrel{\to }{n}$ calculate the area of ​​the triangle created by vectors $\stackrel{\to }{a},\stackrel{\to }{b}$
I found out what $|\stackrel{\to }{a}|$ and $|\stackrel{\to }{b}|$ but not sure how to continue form here.
$\begin{array}{rl}{|\stackrel{\to }{a}|}^{2}& =\left(\stackrel{\to }{m}-\stackrel{\to }{n}\right)\left(\stackrel{\to }{m}-\stackrel{\to }{n}\right)\\ & ={|\stackrel{\to }{m}|}^{2}-2\stackrel{\to }{m}\cdot \stackrel{\to }{n}+{|\stackrel{\to }{n}|}^{2}\\ & =3-2\sqrt{3}\cdot 1\cdot \phantom{\rule{mediummathspace}{0ex}}\frac{\sqrt{3}}{2}+1\\ & =1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\overline{)|\stackrel{\to }{a}|=1}\end{array}$
$\begin{array}{rl}{|\stackrel{\to }{b}|}^{2}& =\left(\stackrel{\to }{m}+\stackrel{\to }{n}\right)\cdot \left(\stackrel{\to }{m}+\stackrel{\to }{n}\right)\\ & ={|\stackrel{\to }{m}|}^{2}+2\stackrel{\to }{m}\cdot \stackrel{\to }{n}+{|\stackrel{\to }{n}|}^{2}\\ & =3+2\sqrt{3}\cdot 1\cdot \phantom{\rule{mediummathspace}{0ex}}\frac{\sqrt{3}}{2}+1\\ & =7\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\overline{)|\stackrel{\to }{b}|=\sqrt{7}}\end{array}$
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abortargy
The area of triangle generated by $\stackrel{\to }{u}$ and $\stackrel{\to }{v}$ is $\frac{1}{2}|\stackrel{\to }{u}×\stackrel{\to }{v}|=\frac{1}{2}|\stackrel{\to }{u}||\stackrel{\to }{v}|\mathrm{sin}\theta$
$\theta$ is 30 degree.
$\stackrel{\to }{a}×\stackrel{\to }{b}=\left(\stackrel{\to }{x}-\stackrel{\to }{y}\right)×\left(\stackrel{\to }{x}+\stackrel{\to }{y}\right)=\stackrel{\to }{x}×\stackrel{\to }{x}+\stackrel{\to }{x}×\stackrel{\to }{y}-\stackrel{\to }{y}×\stackrel{\to }{x}-\stackrel{\to }{y}×\stackrel{\to }{x}=2\stackrel{\to }{x}×\stackrel{\to }{y}$
So the area of triangle generated $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ is twice of area of triangle generated $\stackrel{\to }{x}$ and $\stackrel{\to }{y}$.
So the answer is $|\stackrel{\to }{x}||\stackrel{\to }{y}|\mathrm{sin}{30}^{\circ }=\frac{\sqrt{3}}{2}$