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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.8 question 4 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.8 question 4 maths

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Answer:-  The answer of the given question is \vec{r} \cdot(\hat{\imath}+9 \hat{\jmath}+11 \hat{k})=0

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

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

Solution:- \vec{r} \cdot \overrightarrow{n_{1}}=d_{1}

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

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

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

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

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

As given that, plane (i) is passing through the point 2 \hat{\imath}+\hat{\jmath}-\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 \\\\ (2)(1)+(1)(3)+(-1)(-1)+k[(2)(0)+(1)(1)+(-1)(2)]=0 \end{gathered}

        \begin{gathered} (2+3+1)+k(1-2)=0 \\\\ 6-k=0 \\\\ k=6 \end{gathered}

Putting the value of k in equation (i)

        \begin{gathered} \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{j}+11 \hat{k})=0 \end{gathered}

So, the equation of required plane is

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

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