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  • Provide solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 8

Provide solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 8

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Answer:-  The answer of the given question is \vec{r}\left (\hat{i}+7\hat{j} \right )+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 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+z-1=0

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

x-3y+2z-5=0 and 2x-y+z-1=0 is given by
\left (x-3y+2z-5 \right )+k\left (2x-y+3z-1 \right )=0\\ x(1-2k)+y(-3-k)+z(2+3k)-5-k=0                             … (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

\begin{gathered} x+7 y+13=0 \\ (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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