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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 10

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 10

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Answer:-  The answer of the given question is \vec{r} \cdot(4 \hat{i}+2 \hat{j}-4 \hat{k})+6=0  or

\vec{r} \cdot(-2 \hat{i}+4 \hat{j}+4 \hat{k})+6=0

Hint:-  We know that, equation of a plane passing through the line of intersection of two planes a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0 is given by

\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0


Given:- \begin{aligned} &-\vec{r} \cdot(\hat{\imath}+3 \hat{\jmath})+6=0 \end{aligned}      

            \\\\ \vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})=0

Solution:-  The equation of a plane through the line of intersection of  the planesr \cdot(\hat{\imath}+3 \hat{j})+6=0   and \vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 k)=0

                                    \begin{gathered} \vec{r} \cdot(\hat{\imath}+3 \hat{\jmath})+6+\lambda[\vec{r} \cdot(3 \hat{\imath}-\hat{\jmath}-4 \hat{k})]=0 \\\\ \vec{r} \cdot[(\hat{\imath}+3 \hat{\jmath})]+6+3 \lambda \cdot \vec{r} \hat{\imath}-\vec{r} \lambda \hat{j}-4 \lambda \vec{r} \hat{k}=0 \end{gathered}

                                    r(\hat{\imath}+3 \hat{\jmath}+3 \lambda \hat{\imath}-\lambda \hat{\jmath}-4 \lambda \hat{k})=-6                ...........(i)

                \vec{r} \cdot \frac{[\hat{\imath}(1+3 \lambda)+\hat{j}(3-\lambda)+\hat{k}(-4 \lambda)]}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=\frac{-6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}

The perpendicular distance from the origin is unity

            \begin{aligned} &\frac{-6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=1 \\\\ &(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}=36 \end{aligned}

            \begin{gathered} 1+9 \lambda^{2}+6 \lambda+9+\lambda^{2}-6 \lambda+16 \lambda^{2}=36 \\\\ \lambda^{2}=1 \\\\ \lambda=\pm 1 \end{gathered}

Using equation (i) the required plane is

                \vec{r} \cdot(1 \pm 3) \hat{\imath}+(3 \mp 1) \hat{\jmath}+(\mp 4) \hat{k}=-6

                \vec{r} \cdot(4 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})=-6 \text { or }\vec{r} \cdot(-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})=-6

                \vec{r} \cdot(4 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})+6=0 \text { orr. }(-2 \hat{\imath}+4 \hat{\jmath}+4 \hat{k})+6=0

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