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

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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