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If the line of intersection of the planes \mathrm{a x+b y=3 \: and \: a x+b y+c z=0, a >0} makes an angle \mathrm{30^{\circ}} with the plane \mathrm{y-z+2=0}, then the direction cosines of the line are :
 

Option: 1

\mathrm{\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0}
 


Option: 2

\mathrm{ \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, 0 }


Option: 3

\mathrm{\frac{1}{\sqrt{5}},-\frac{2}{\sqrt{5}}, 0}

 


Option: 4

\mathrm{\frac{1}{2},-\frac{\sqrt{3}}{2}, 0}
 


Answers (1)

best_answer

\begin{aligned} &\text{Direction ratios of the line of intersection}\\ &=\left|\begin{array}{ccc} \mathrm{i }& \mathrm{j} & \mathrm{k} \\ \mathrm{a} & \mathrm{b} & 0 \\ \mathrm{a} & \mathrm{b} & \mathrm{c} \end{array}\right| \\ &=\mathrm{b c i-a c j} \\ &=\mathrm{c(b i-a j)}\\ &\therefore \text{Direction ratios of b,-a, 0}\\ &\text{Angle between line and }\mathrm{ y-z+2=0 }\text{ is }\mathrm{30^{\circ}} \\ &\therefore\mathrm{ \sin \left(30^{\circ}\right)=\left|\frac{(b i-a j) \cdot(j-k)}{\sqrt{a^{2}+b^{2}} \sqrt{1+1}}\right| }\\ &\Rightarrow \quad \mathrm{\frac{1}{2}=\left|\frac{-a}{\sqrt{a^{2}+b^{2}} \sqrt{2}}\right| }\\ &\Rightarrow \mathrm{\sqrt{a^{2}+b^{2}}=\sqrt{2}|a| }\\ &\Rightarrow \mathrm{a^{2}+b^{2}=2 a^{2} }\\ &\Rightarrow \mathrm{a^{2}=b^{2}}\\ &\text{option (A) and (B) satisfy these conditions} \end{aligned}

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manish painkra

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