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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

Answers (1)

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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