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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 19

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 19

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Answer:-  The answer of the given question is x+2y+3z=4.

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:-x+y+z=1

            2x+3y+4z=5     

∴ Required equation of plane is x+y-1+z+\lambda(2 x+3 y+4 z-5)=0 for some λ

            i.e.(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z=(1+5 \lambda)

According to question

            2\left(\frac{1+5 \lambda}{1+3 \lambda}\right)=3\left(\frac{1+5 \lambda}{1+4 \lambda}\right)

Solving we get \lambda =-1

Thus the equation of required plane is

            -x-2y-3z=-4

            x+2y+3z=4

Posted by

infoexpert26

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