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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

Answers (1)

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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