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  • Explain solution RD sharma class 12 chapter 28 The Plane exercise 28.8, question 19

Explain solution RD sharma class 12 chapter 28 The Plane exercise 28.8, question 19

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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 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 (2x+3y+4z-5)=0  for some \lambda

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

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