A) \[\frac{5}{3}\]
B) \[\frac{2}{3}\]
C) \[\frac{3}{2}\]
D) \[\frac{2}{5}\]
Correct Answer: B
Solution :
The direction ratios of the given line \[\frac{x}{1}=\frac{y-1}{2}=\frac{z-3}{\lambda }\] are 1, 2, \[\lambda \] Hence direction cosines of the line are\[\frac{1}{\sqrt{5+{{\lambda }^{2}}}},\frac{2}{\sqrt{5+{{\lambda }^{2}}}},\frac{\lambda }{\sqrt{5+{{\lambda }^{2}}}}\] Also the direction cosines of the normal to the plane are \[\frac{1}{\sqrt{14}}m\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\]. Angle between the line and the plane is \[\theta \], then \[\cos ({{90}^{o}}-\theta )=\frac{1+4+3\lambda }{\sqrt{\left( 5+{{\lambda }^{2}} \right)\sqrt{14}}}=\sin \theta \] \[\Rightarrow \,\,\,\frac{3}{\sqrt{14}}=\frac{5+3\lambda }{\sqrt{{{\lambda }^{2}}+5}\sqrt{14}}\] \[\Rightarrow \,\,9{{\lambda }^{2}}+45=9{{\lambda }^{2}}+30\lambda +25\] \[\Rightarrow 30\lambda =20\] \[\lambda =\frac{2}{3}\]
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