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Question

Find the equation of the plane passing through (2,2,2) and (2,2,2) and perpendicular to the plane 2x3y+z7=0.

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Solution

Suppose required plane is ax+by+ca+d=0
¯n=(a,b,c) is normal to the plane
Required plane is perpendicular to 2x3y+z7=0
¯n(2,3,1)=0 (1)
Now A(2,2,2) and B(2,2,2) are the points of plane.
AB=BA=(4,4,4)
¯nAB=0
¯n(4,4,4)=0
¯n(1,1,1)=0 .(2)

From result (1) and (2) we have
¯n=(2,3,1)×(1,1,1)
=∣ ∣ijk231111∣ ∣
=i(3+1)j(2+1)+k(2,2)
=2^i3^j5^k
¯n=(2,3,5) OR
¯n=(2,3,5)

If plane passes from point A(¯a)=(2,2,1)
Then we get its equation using ¯r¯n=¯a¯n
(x,y,z)(2,3,5)=(2,2,2)(2,3,5)
2x+3y+5z=4+610
2x+3y+5z=0
Which is required equation of plane.

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