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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.8 question 8 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.8 question 8 maths

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Answer:-  The answer of the given question is \vec{r}(\hat{\imath}+7 \hat{\jmath})+13=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:-x-3y+2z-5=0 and 2x-y+3z-1=0

Solution:-  So equation of plane passing through the line of intersection of planes

x-3y+2z-5=0 and 2x-y+3z-1=0 is given by
\begin{aligned} (x-3 y+2 z-5) &+k(2 x-y+3 z-1) &=0 & \\\\ x(1-2 k)+y(-3-k)+z(2+3 k)-5-k &=0 & \end{aligned} .     … (i)                       

Plane (i) is passing the through the point(1,-2,3), so


\begin{gathered} 1(1+2 k)+(-2)(-3-k)+(3)(2+3 k)-5-k=0 \\\\ 1+2 k+6+2 k+6+9 k-5-k=0 \\\\ 8+12 k=0 \\\\ k=-\frac{2}{3} \end{gathered}

Putting the value of k in equation (i)

                            \begin{gathered} x(1+2 k)+y(-3-k)+z(2+3 k)-5-k=0 \\\\ x\left(1-\frac{4}{3}\right)+y\left(-3+\frac{2}{3}\right)+z\left(2-\frac{6}{3}\right)-\left(\frac{15-2}{3}\right)=0 \\\\ -\frac{1}{3} x-\frac{7}{3} y-\frac{13}{3}=0 \end{gathered}

Multiplying by -3

                                    x+7y+13=0
                        \begin{gathered} (x \hat{\imath}+y \hat{\jmath}+z \hat{k})(\hat{\imath}+7 \hat{\jmath})+13=0 \\\\ \vec{r} \cdot(\hat{\imath}+7 \hat{\jmath})+13=0 \end{gathered}

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