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Find the equation of plane passing through(2,2,1) and (9,3,6) and perpendicular to the plane x+3y+3z=8.

Find the equation of plane passing through(2,2,1) and (9,3,6) and perpendicular to the plane x+3y+3z=8.

Grade:12th Pass

1 Answers

Rinkoo Gupta
askIITians Faculty 80 Points
7 years ago
Eq of the plan epassing through (2,2,1)is A(x-2)+B(y-2)+C(z-1)=0 ..........EQ (1)
It passes through (9,3,6) So this will satisfy eq of plane
so A(9-2)+B(3-2)+C(6-1)=0
=>7A+B+5C=0 ...........EQ(2)
the eq (1) is perpendic ular to the plane x+3y+3z=8
So A+3B+3C=0 ............EQ(3)
Solving eq(2) and (3) by cross multiplication method
A/(-12)=B/(-16)=C/20 (=k ,let)
then A=-12k,B=-16k and C=20k
Putting these values in eq(1), we get
-12k(x-2)-16k(y-2)+20k(z-1)=0
=> -12(x-2)-16(y-2)+20(z-1)=0
=>-12x+24-16y+32+20z-20=0
=>-12x-16y+20z+36=0
=>3x+4y-5z-9=0

Thanks & Regards
Rinkoo Gupta
AskIItians Faculty

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