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  • Need solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 12

Need solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 12

Answers (1)

 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 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}+2 \hat{\jmath}+3 \hat{k})-4=0 \\ \end{aligned} ........(i)

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

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

                           
\begin{aligned} &{[\vec{r} .(\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 \\ \end{aligned}     ............(iii)

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

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

We know that, two planes are perpendicular if  \begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\end{aligned}

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

 

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

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

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

This is vector equation of the required plane

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

\begin{aligned} &33 x+45 y+50 z-41=0 \end{aligned}

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