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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.8 question 9 maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.8 question 9 maths textbook solution

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Answer:-  The answer of the given question is 51x+15y-50z+173=0.

Hints:-  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:-  5x+3y+6z+8=0                            

(x +2y+3z-4)=0        and

x +2y+3z-4=0 

Solution:-  The equation of a plane through the line of intersection of  the planes x+2y+3z-4=0 and 2x+y-z+5=0

                \begin{aligned} (x+2 y+3 z-4) &+\lambda(2 x+y-z+5)=0 \\\\ x(1+2 \lambda)+y(2+\lambda)+z(-\lambda+3)-4+5 \lambda &=0 \end{aligned} .  … (i)


Also this is perpendicular to the plane 5x+3y+6z+8=0   …(ii)

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


                                    \begin{gathered} 5(1+2 \lambda)+3(2+\lambda)+6(-\lambda+3)=0 \\\\ \therefore 5+10 \lambda+6+3 \lambda+18-6 \lambda=0 \\\\ \lambda=-\frac{29}{7} \end{gathered}

∴ Putting this value of \lambda   in equation (i) we get equation of plane as

                                            51x+15y-50z+173=0

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