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

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

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Answer:-  The answer of the given question is 33x+45y+50z-41=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:-  \vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})-4=0        … (i)

             \vec{r} \cdot(2 \hat{\imath}+\hat{\jmath}-\hat{k})+5=0           … (ii)

Solution:-  The equation of the plane passing through the line of intersection of the plane given in equation (i) and equation (ii) is

            [\vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})-4]+\lambda[\vec{r} \cdot(2 \hat{\imath}+\hat{\jmath}-\hat{k})+5]=0

            \vec{r}[(2 \lambda+1) \hat{\imath}+(\lambda+2) \hat{\jmath}+(3-\lambda) \hat{k}]+(5 \lambda-4)=0           …(iii)

The plane in equation (iii) is perpendicular to the plane,

            \vec{r} \cdot(5 \hat{\imath}+3 \hat{\jmath}-6 \hat{k})+8=0

We know that, two planes are perpendicular if  a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0


        \begin{gathered} \therefore 5(2 \lambda+1)+3(\lambda+2)-6(3-\lambda)=0 \\\\ 19 \lambda-7=0 \\\\ \lambda=\frac{7}{19} \end{gathered}

Substituting \lambda=\frac{7}{19}  in equation (iii), we obtain

        \vec{r} \cdot\left[\frac{33}{19} \hat{\imath}+\frac{45}{19} \hat{\jmath}+\frac{50}{19} \hat{k}\right]-\frac{41}{19}=0

        \vec{r} \cdot(33 \hat{\imath}+45 \hat{\jmath}+50 \hat{k})-41=0              … (iv)

This is vector equation of the required plane

Now (x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(33 \hat{\imath}+45 \hat{\jmath}+50 \hat{k})-41=0

        33x+45y+50z-41=0

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