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  • Provide Solution for RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 18

Provide Solution for RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 18

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Answer: Therefore, required equation of plane is  \vec{r} \cdot(-9 \hat{i}+8 \hat{j}-\hat{k})=11 \text { and }-9 x+8 y-z=11  and distance of plane is \sqrt{146} \: units

Hint: Use properties of plane

Given: \begin{aligned} &\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k}) \\ & \end{aligned}

            \vec{r}=(\hat{i}-3 \hat{j}+5 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})

Solution: The plane passes through the point \left ( 1,2,-4 \right )

            A vector in a direction perpendicular to \vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+m(2 \hat{i}+3 \hat{j}+6 \hat{k})  and

            \overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})+\mathrm{n}(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \text { is } \overrightarrow{\mathrm{n}}=(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \times(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})

            \Rightarrow \vec{n}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{array}\right|=-9 \hat{i}+8 \hat{\mathrm{j}}-\hat{k}

            Equation of the plane is  (\vec{r}-\vec{a}) \cdot \vec{n}=0

            \begin{aligned} &\{\vec{r}-(\hat{i}+2 \hat{j}-4 \hat{k})\} \cdot(-9 \hat{i}+8 \hat{j}-\hat{k})=0 \\ & \end{aligned}

             \Rightarrow \vec{r} \cdot(-9 \hat{i}+8 \hat{j}-\hat{k})=11

Substitution \vec{r}=x \hat{i}+y \hat{j}+z \hat{k}  , we get the Cartesian from as  -9 x+8 y-z=11

            The distance of the point (9,8,-10)  from the plane

                        \begin{aligned} &=\frac{-9 \times 9+8 \times(-8)-1 \times(-10)-11}{\sqrt{-9^{2}+8^{2}+-1^{2}}} \\ & \end{aligned}

                        =\frac{-81-64+10-11}{\sqrt{81+64+1}} \\

                        =\left|\frac{-146}{\sqrt{146}}\right|=\sqrt{146} \text { units }

 

Posted by

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