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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 14

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 14

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

Hints:- We know the line of intersection of the plane \vec{r} \cdot \vec{n}_{1}-\vec{d}_{1}=0 \text { and } \vec{r} \cdot \vec{n}_{1}-\vec{d}_{2}=0  is

            given by  \vec{r} \cdot\left(\vec{n}_{1}+k \vec{n}_{2}\right)-d_{1}+k d_{2}=0

Given :-The intersection of the plane vector

            \begin{aligned} &\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})=7 \quad \text { and } \\\\ &\vec{r} \cdot(2 \hat{i}+3 \hat{j}+3 \hat{k})=9 \end{aligned}

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

            \vec{r} \cdot \vec{n}_{1}-d_{1}=0 \text { and } \vec{r} \cdot \vec{n}_{1}-d_{2}=0

Is given by  \vec{r} \cdot\left(\vec{n}_{1}+k \vec{n}_{2}\right)-d_{1}+k d_{2}=0

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

            \begin{aligned} &\vec{r} \cdot(2 \hat{i}+\hat{j}+3 \hat{k})-7=0 \text { and } \\ &\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})-9=0 \end{aligned}is given by

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

Given that plane (1) is passing through  2 \hat{i}+\hat{j}+3 \hat{k} \text { 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 \\\\ &(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 \\\\ &\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 \\\\ &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} \cdot(2 \hat{i}-13 \hat{j}+3 \hat{k})=0

Equation of the required plane is

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

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