question_answer

If the plane \[2x-3y+6z-11=0\]makes an angle \[\theta \] with the X-axis, then the value of \[\tan \theta \]is equal to

A)  \[\frac{2}{3}\]

B)  \[\frac{2}{15}\]

C)  \[\frac{\sqrt{2}}{3}\]

D)  \[\frac{2\sqrt{5}}{15}\]

Correct Answer: D

Solution :

Let the equation of X-axis be \[\frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0}\] Here, \[{{a}_{2}}=a,\,\,\,{{b}_{1}}=0\]  and \[{{c}_{1}}=0\] Given equation of plane is \[2x-3y+6z-11=0\] Here, \[{{a}_{2}}=2,\,\,{{b}_{2}}=-3,\,{{c}_{2}}=6\] Let \[\theta \] be the angle between line and plane. Then, \[\sin \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\,\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\] \[=\frac{a\times 2+0\times (-3)+0\times 6}{\sqrt{{{a}^{2}}+{{0}^{2}}+{{0}^{2}}}\,\sqrt{{{2}^{2}}+{{(-3)}^{2}}+{{6}^{2}}}}\] \[=\frac{2a}{\sqrt{{{a}^{2}}}\sqrt{4+9+36}}=\frac{2a}{a.\,7}=\frac{2}{7}\] \[\Rightarrow \] \[\sin \,\,\theta =\frac{2}{7}\] \[\therefore \] \[\tan \,\,\theta =\frac{2}{\sqrt{45}}\] \[\Rightarrow \] \[\tan \,\theta =\frac{2}{3\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}\] \[\Rightarrow \] \[\tan \,\theta =\frac{2\sqrt{5}}{15}\]