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Question Number 112902 by mohammad17 last updated on 10/Sep/20

$$findtheanglebetween3y+\frac{x}{\sqrt{3}}=1,\frac{\sqrt{3}}{2}y-x=2$$$$helpmesir$$

Answered by bemath last updated on 10/Sep/20

$${\mathrm{m}}_1=\frac{-\frac{1}{\sqrt{3}}}{3}=-\frac{1}{3\sqrt{3}};{\mathrm{m}}_2=\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}=\frac{2}{\sqrt{3}}$$$$\mathrm{say}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{both}\mathrm{line}$$$$\mathrm{is}\alpha \mathrm{then}\tan \alpha =\mid \frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{1+{\mathrm{m}}_2.{\mathrm{m}}_1}\mid$$$$\tan \alpha =\mid \frac{\frac{6+1}{3\sqrt{3}}}{1-\frac{2}{9}}\mid =\frac{\frac{21}{\sqrt{3}}}{7}=\frac{3}{\sqrt{3}}$$$$\alpha ={\tan }^{-1}\left(\sqrt{3}\right)=60°$$

Commented by mohammad17 last updated on 10/Sep/20

$$thankyousir$$

Answered by mathmax by abdo last updated on 10/Sep/20

$$3\mathrm{y}+\frac{\mathrm{x}}{\sqrt{3}}=1\Rightarrow 3\sqrt{3}\mathrm{y}+\mathrm{x}=\sqrt{3}\Rightarrow \mathrm{x}+3\sqrt{3}\mathrm{y}-\sqrt{3}=0$$$$\Rightarrow \mathrm{the}\mathrm{director}\mathrm{verctor}\mathrm{is}\overrightarrow{\mathrm{u}}(-3\sqrt{3},1)$$$$\frac{\sqrt{3}}{2}\mathrm{y}-\mathrm{x}=2\Rightarrow \sqrt{3}\mathrm{y}-2\mathrm{x}=4\Rightarrow -2\mathrm{x}+\sqrt{3}\mathrm{y}-4=0\Rightarrow 2\mathrm{x}-\sqrt{3}\mathrm{y}+4=0\Rightarrow$$$$\mathrm{the}\mathrm{director}\mathrm{vector}\mathrm{is}\overrightarrow{\mathrm{v}}(\sqrt{3},2)$$$$\cos (\mathrm{u},\mathrm{v})=\frac{\mathrm{u}.\mathrm{v}}{\mid \mathrm{u}\mid .\mid \mathrm{v}\mid }\mathrm{and}\sin (\mathrm{u},\mathrm{v})=\frac{\det (\mathrm{u},\mathrm{v})}{\mid \mathrm{u}\mid .\mid \mathrm{v}\mid }\Rightarrow \tan (\mathrm{u},\mathrm{v})=\frac{\det (\mathrm{u},\mathrm{v})}{\mathrm{u}.\mathrm{v}}$$$$=\frac{\left|\begin{array}{c}-3\sqrt{3}\sqrt{3}\\ 12\end{array}\right|}{(-3\sqrt{3}).\sqrt{3}+1\times 2}=\frac{-7\sqrt{3}}{-7}=\sqrt{3}$$$$\mathrm{we}\mathrm{have}\tan (\mathrm{u},\mathrm{v})=\sqrt{3}\Rightarrow (\mathrm{u},\mathrm{v})=(\mathrm{u},\mathrm{v})=\frac{\pi }{3}\left[\pi \right]$$

Commented by mohammad17 last updated on 10/Sep/20

$$thankyousir$$

Commented by mathmax by abdo last updated on 11/Sep/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

Answered by Dwaipayan Shikari last updated on 10/Sep/20

$$3y+\frac{x}{\sqrt{3}}=1\frac{\sqrt{3}}{2}y-x=2\Rightarrow y=\frac{2}{\sqrt{3}}x+\frac{4}{\sqrt{3}}$$$$y=-\frac{x}{3\sqrt{3}}+\frac{1}{3}$$$$tan\theta =\mid \frac{-\frac{1}{3\sqrt{3}}-\frac{2}{\sqrt{3}}}{1-\frac{2}{9}}\mid =\sqrt{3}$$$$\theta =\frac{\pi }{3}$$$$$$

Commented by mohammad17 last updated on 10/Sep/20

$$thankyousir$$

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