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  • Please Solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 1 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 1 Maths Textbook Solution.

Answers (1)

Answer:  \sin ^{-1} \frac{9}{\sqrt{89}}

Hint: Use formula  \vec{r}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+9 \hat{k})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{k}) \text { and } \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=5

Given:

Solution: Equation of line is   \vec{r}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+9 \hat{k})+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{k})   and the equation of plane is \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=5

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

\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}}}

Here, a_{1}=1, b_{1}=1, c_{1}=1

and  a_{2}=2, b_{2}=3, c_{2}=4

The angle between them is

            \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}}}

                      \begin{aligned} &=\frac{1 \times 2+1 \times 3+1 \times 4}{\sqrt{1^{2}+1^{2}+1^{2}} \sqrt{2^{2}+3^{2}+4^{2}}} \\ & \end{aligned}

                      =\frac{2+3+4}{\sqrt{3} \sqrt{29}} \\

                     =\frac{9}{\sqrt{87}} \\

            \theta=\sin ^{-1} \frac{9}{\sqrt{87}}

Posted by

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