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Please Solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 22 Maths Textbook Solution.

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Answer:  \sin ^{-1}\left(\frac{8}{21}\right)

Hint: Use formula  \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}}}

Given:  \frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6} \text { and } 10 x+2 y-11 z=3

Solution: Direction ratios of the line

            \frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6} \text { are }(2,3,6)

Direction ratio of a line perpendicular to the plane  10 x+2 y-11 z=3 \text { are }(10,2,-11)

As we know that the angle \theta  between the line \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}  and plane

a_{2} x+b_{2} y+c_{2} z+d_{2}=0  is given by

            \begin{aligned} &\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}}} \\ & \end{aligned}

            \sin \theta=\frac{2 \times 10+3 \times 2+6 \times(-11)}{\sqrt{2^{2}+3^{2}+6^{2}} \sqrt{10^{2}+2^{2}+(-11)^{2}}}

            \begin{aligned} \sin \theta &=\frac{-40}{\sqrt{49} \sqrt{225}} \\ & \end{aligned}

                     =\frac{-40}{7 \times 15}=\frac{-8}{21}

           \begin{gathered} \sin \theta=\frac{-8}{21} \\ \end{gathered}

                \theta=\sin ^{-1}\left(\frac{8}{21}\right)

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