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  • Please Solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 3 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 3 Maths Textbook Solution.

Answers (1)

Answer:  \sin ^{-1}\left(\frac{23}{11 \sqrt{11}}\right)

Hint: Use formula  \sin \theta=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}

Given:  (3,-4,-2) \&(12,2,0) \text { and plane } 3 x-y+z=1

Solution: It is given that the line passes through  (3,-4,-2) \&(12,2,0)

So, \begin{aligned} \vec{b}=\overrightarrow{\mathrm{AB}} &=\overline{\mathrm{OB}}-\overline{\mathrm{OA}} \\ & \end{aligned}

                      =12 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}-(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \\

                      =9 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}

The given line is parallel to the vector \overrightarrow{\mathrm{b}}=9 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} and given line is normal to the vector  \vec{n}=3 \hat{i}-\hat{j}+\hat{k}

We know that, the angle  \theta  between the line and plane is given by

            \begin{aligned} \sin \theta &=\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|} \\ & \end{aligned}

                      =\frac{(9 \hat{i}+6 \hat{j}+2 \hat{k}) \cdot(3 \hat{i}-\hat{j}+\hat{k})}{|9 \hat{i}+6 \hat{j}+2 \hat{k} \| 3 \hat{i}-\hat{j}+\hat{k}|} \\

                      =\frac{27-6+2}{\sqrt{81+36+4} \sqrt{9+1+1}}=0 \\

                      =\frac{23}{11 \sqrt{11}} \\

            \Rightarrow \theta =\sin ^{-1} \frac{23}{11 \sqrt{11}}

 

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