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  • Provide solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 9

Provide solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 9

Answers (1)

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  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

2x+y-z+5=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  
\left (x+2y+3z-4 \right )+\lambda \left (2x+y-z+5 \right )=0\\ x(1+2\lambda)+y(2+\lambda)+z(-\lambda+3)-4+5\lambda=0                              … (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

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}
∴ Putting this value of λ in equation (i) we get equation of plane as

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

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