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  • Provide solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 10

Provide solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 10

Answers (1)

The answer of the given question is \vec{ r}.\left (4\hat{i}+2\hat{j}-4\hat{k} \right )+6=0  or  \vec{ r}.\left (-2\hat{i}+4\hat{j}+4\hat{k} \right )+6=0

Hint:-  We know that, equation of a plane passing through the line of intersection of two planes

 a_{1}x+b_{1}y+c_{1}z+d_{1}=0 and a_{2}x+b_{2}y+c_{2}z+d_{2}=0  is given by

(a_{1}x+b_{1}y+c_{1}z+d_{1})+k(a_{2}x+b_{2}y+c_{2}z+d_{2})=0

Given:-  \vec{r}.(\hat{i}+3\hat{j}+6)=0                                          

                \vec{r}.(3\hat{i}-\hat{j}+-4\hat{k})=0                      

Solution:-  The equation of a plane through the line of intersection of  the planes  \vec{r}.(\hat{i}+3\hat{j}+6)=0     and   \vec{r}.(3\hat{i}-\hat{j}+-4\hat{k})=0                      

\begin{gathered} \vec{r} \cdot(\hat{\imath}+3 \hat{\jmath})+6+\lambda[\vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})]=0 \\ \end{gathered}

\begin{gathered} \vec{r} \cdot[(\hat{\imath}+3 \hat{\jmath})]+6+3 \lambda \cdot \vec{r} \hat{\imath}-\vec{r} \lambda \hat{\jmath}-4 \lambda \vec{r} \hat{k}=0 \\ \end{gathered}

\begin{gathered} r(\hat{\imath}+3 \hat{\jmath}+3 \lambda \hat{\imath}-\lambda \hat{\jmath}-4 \lambda \hat{k})=-6 \\ \end{gathered}                                 ..........(i)\begin{gathered} \vec{r} \cdot \frac{[\hat{\imath}(1+3 \lambda)+\hat{\jmath}(3-\lambda)+\hat{k}(-4 \lambda)]}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=\frac{-6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}} \\ \end{gathered}

The perpendicular distance from the origin is unity

\begin{gathered} \frac{-6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=1 \\ \end{gathered}

\begin{gathered} \begin{array}{c} (1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}=36 \\ 1+9 \lambda^{2}+6 \lambda+9+\lambda^{2}-6 \lambda+16 \lambda^{2}=36 \\ \lambda^{2}=1 \\ \lambda=\pm 1 \end{array} \end{gathered}

Using equation (i) the required plane is
\vec{r}.(1\pm 3)\hat{i}+(3\mp 1)\hat{j}+(\mp 4)\hat{k}=-6\\ \vec{r}(4\hat{i}+2\hat{j}-4\hat{k})=-6\;\;\; or \;\;\; \vec{r}(-2\hat{i}+4\hat{j}+4\hat{k})=-6\\ \vec{r}(4\hat{i}+2\hat{j}-4\hat{k})+6=0\;\;\; or \;\;\; \vec{r}(-2\hat{i}+4\hat{j}+4\hat{k})+6=0\\

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