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Need solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 14

Answers (1)

The answer of the given question is \vec{r}.\left ( 2\hat{i}-13\hat{j}+3\hat{k} \right )=0

Hints:- We know the line of intersection of the plane \vec{r}.\vec{n_{1}}-\vec {d}_{1}=0 and \vec{r}.\vec{n_{1}}-\vec {d}_{2}=0 is given by

               \vec{r}.\left (\vec{n_{1}}+k\vec{n_{2}} \right )-\vec {d}_{1}+k\vec {d}_{2}=0

Given :-The intersection of the plane vector

               \vec{r}.\left ( 2\hat{i}+\hat{j}+3\hat{k} \right )=7  and

  \vec{r}.\left ( 2\hat{i}+3\hat{j}+3\hat{k} \right )=9

Solution:-We know that, the equation of a plane through the line of intersection of the plane

               \vec{r}.\vec{n_{1}}-\vec {d}_{1}=0 and \vec{r}.\vec{n_{1}}-\vec {d}_{2}=0

Is given by   \vec{r}.\left (\vec{n_{1}}+k\vec{n_{2}} \right )-\vec {d}_{1}+k\vec {d}_{2}=0

So, equation of plane passing through the line of intersection of plane

   \vec{r}.\left ( 2\hat{i}+\hat{j}+3\hat{k} \right )-7=0  and

  \vec{r}.\left ( 2\hat{i}+3\hat{j}+3\hat{k} \right )-9=0  is given by

 \left [\vec{r}.\left ( 2\hat{i}+\hat{j}+3\hat{k} \right )-7 \right ] +k\left [\vec{r}.\left ( 2\hat{i}+3\hat{j}+3\hat{k} \right )-9 \right ]=0 \\

\vec{r} \left [ (2+2k)\hat{i}+(1+5k)\hat{j}+(3+3k)\hat{k} \right ] -7-9k=0                                      --------(1)

Given that plane (1) is passing through 2\hat{i}+\hat{j}+3\hat{k} so

             \begin{aligned} &(2 \hat{i}+\hat{j}+3 \hat{k})[(2+2 k) \hat{i}+(1+5 k) \hat{j}+(3+3 k) \hat{k}]-7-9 k=0 \\ \end{aligned}

\begin{aligned} &(2)(2+2 k)+(1)(1+5 k)+(3)(3+3 k)-7-9 k=0 \\ &4+4 k+1+5 k+9+9 k-7-9 k=0 \\ \end{aligned}

\begin{aligned} &9 k=-7 \\ &k=\frac{-7}{9} \\ \end{aligned}

Put the value of k in equation (1)

             \begin{aligned} &\vec{r}[(2+2 k) \hat{i}+(1+5 k) \hat{j}+(3+3 k) \hat{k}]-7-9 k=0 \\ \end{aligned}

\begin{aligned} &\vec{r}\left[\left(2-\frac{14}{9}\right) \hat{i}+\left(1-\frac{35}{9}\right) \hat{j}+\left(3-\frac{21}{9}\right) \hat{k}\right]-7+\frac{63}{9}=0 \\ \end{aligned}

\begin{aligned} &\vec{r}\left[\left(\frac{18-14}{9}\right) \hat{i}+\left(\frac{9-35}{9}\right) \hat{j}+\left(\frac{27-21}{9}\right) \hat{k}\right]-7+7=0 \\ \end{aligned}

\begin{aligned} &r\left[\left(\frac{4}{9}\right) \hat{i}-\left(\frac{26}{9}\right) \hat{j}+\left(\frac{6}{9}\right) \hat{k}\right]=0 \end{aligned}

Multiplying by \frac{9}{2},we get

            \vec{r}.\left ( 2\hat{i}-13\hat{j}+3\hat{k} \right )=0

Equation of the required plane is

               \vec{r}.\left ( 2\hat{i}-13\hat{j}+3\hat{k} \right )=0

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