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The plane \( 2 x-3 y+6 z-11=0 \) makes an angle \( \sin ^{-1}(\alpha) \) with \( X \)-axis. The value of a is equal
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Verified Answer
The correct answer is:
\( \frac{2}{7} \)
Given plane,
\( u: 2 x-3 y+6 z-11=0 \rightarrow(1) \)
Equation of plane on x-axis is given by
\( v: 1 x+0 y+0 z=0 \rightarrow(2) \)
Angle between two planes is given by
\[
\begin{array}{l}
\sin \theta=\frac{|u \cdot v|}{|u| \cdot|v|} \\
\Rightarrow \sin \theta=\frac{|(2) \cdot(1)+(-3) \cdot(0)+(6) \cdot(0)|}{\sqrt{2^{2}+3^{2}+6^{2}} \cdot \sqrt{1^{2}}} \\
\Rightarrow \sin \theta=\frac{2}{\sqrt{49}}=\frac{2}{7} \Rightarrow \theta=\sin ^{-1}\left(\frac{2}{7}\right)
\end{array}
\]
Since, \( \sin ^{-1} \alpha=\theta \). Then,
\( \sin ^{-1} \alpha=\sin ^{-1}\left(\frac{2}{7}\right) \Rightarrow \alpha=\frac{2}{7} \)
\( u: 2 x-3 y+6 z-11=0 \rightarrow(1) \)
Equation of plane on x-axis is given by
\( v: 1 x+0 y+0 z=0 \rightarrow(2) \)
Angle between two planes is given by
\[
\begin{array}{l}
\sin \theta=\frac{|u \cdot v|}{|u| \cdot|v|} \\
\Rightarrow \sin \theta=\frac{|(2) \cdot(1)+(-3) \cdot(0)+(6) \cdot(0)|}{\sqrt{2^{2}+3^{2}+6^{2}} \cdot \sqrt{1^{2}}} \\
\Rightarrow \sin \theta=\frac{2}{\sqrt{49}}=\frac{2}{7} \Rightarrow \theta=\sin ^{-1}\left(\frac{2}{7}\right)
\end{array}
\]
Since, \( \sin ^{-1} \alpha=\theta \). Then,
\( \sin ^{-1} \alpha=\sin ^{-1}\left(\frac{2}{7}\right) \Rightarrow \alpha=\frac{2}{7} \)
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