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  • please solve RD sharma class 12 chapter 28 The Plane exercise 28.8 question 4 maths textbook solution

please solve RD sharma class 12 chapter 28 The Plane exercise 28.8 question 4 maths textbook solution

Answers (1)

The answer of the given question is \vec{r}.(\hat{i}+9\hat{j}+11\hat{k})=0

Hint:- \vec{r}.\left (\vec{n_{1}}+k\vec{n_{2}} \right )=d_{1}+kd_{2}

Given:- \vec{r}.(\hat{i}+3\hat{j}-\hat{k})=0 and \vec{r}.(\hat{j}+2\hat{k})=0 , Point (2\hat{i}+\hat{j}-\hat{k})=0

Solution:-  \vec{r}.\vec{n_{1}}=d_{1}

\vec{r}.\vec{n_{2}}=d_{2}is given by

\vec{r}.\left (\vec{n_{1}}+k\vec{n_{2}} \right )=d_{1}+kd_{2}

∴ The equation of the plane passing through the line of intersection of given two planes

\vec{r}.(\hat{i}+3\hat{j}-\hat{k})=0 and \vec{r}.(\hat{j}+2\hat{k})=0 is given by

\vec{r}.\left \{ (\hat{i}+3\hat{j}-\hat{k})+k(\hat{j}+2\hat{k}) \right \}=0                                     … (i)

As given that, plane (i) is passing through the point 2\hat{i}+\hat{j}-\hat{k}
\begin{gathered} (2 \hat{\imath}+\hat{\jmath}-\hat{k})(\hat{\imath}+3 \hat{\jmath}-\hat{k})+k(2 \hat{\imath}+\hat{\jmath}-\hat{k})(\hat{\jmath}+2 \hat{k})=0 \\ \end{gathered}

\begin{gathered}(2)(1)+(1)(3)+(-1)(-1)+k[(2)(0)+(1)(1)+(-1)(2)]=0 \\ (2+3+1)+k(1-2)=0 \\ 6-k=0 \\ k=6 \\ \end{gathered}
Putting the value of k in equation (i)

\qquad \begin{array}{c} \vec{r} \cdot\{(\hat{\imath}+3 \hat{\jmath}-\hat{k})+k(\hat{\jmath}+2 \hat{k})\}=0 \\ \vec{r} \cdot\{(\hat{\imath}+3 \hat{\jmath}-\hat{k})+6(\hat{\jmath}+2 \hat{k})\}=0 \\ \vec{r} \cdot(\hat{\imath}+9 \hat{\jmath}+11 \hat{k})=0 \end{array}

So, the equation of required plane is

\vec{r} \cdot(\hat{\imath}+9 \hat{\jmath}+11 \hat{k})=0

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