Question 62453: Let P1 be the plane defined by the points:
(-1,-2,-3), (-3,1,-5) and (2,-5,-4)
Find the equation of the plane perpendicular to P1 and containing the line:
(x,y,z) = (2,-2,-3) + t (2,-3,0)
Your answer should be in the form Ax + By + Cz - D = 0.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! Let P1 be the plane defined by the points:
Q=(-1,-2,-3), R=(-3,1,-5) and S=(2,-5,-4)
Find the equation of the plane perpendicular to P1 and containing the line:
L1 SAY =(x,y,z) = (2,-2,-3) + t (2,-3,0)
EQN. OF REQUIRED PLANE IS
A(X-2)+B(Y+2)+C(Z+3)=0..............1..WITH DRS OF A,B,C.
DRS OF L1 ARE 2,-3,0...IT IS PERPENDICULAR TO NORMAL O REQD.PLANE.HENCE
2A-3B=0.......A=1.5B.........2
DRS OF NORMAL TO PLANE P1 IS GIVEN BY CROSS PRODUCT OF JOINS OF QR AND RS
DRS OF QR = [-3+1,1+2,-5+3]= [-2,3,-2]
DRS OF RS = [2+3,-5-1,-4+5]= [5,-6,1]
QR X RS = [-2i+3j-2k]X[5i-6j+k]= 12k+2j-15k+3i-10j-12i=-9i-8j-3k
THIS SHOULD BE PERPENDICULAR TO NORMAL OF REQD.PLAN.HENCE
-9A-8B-3C=0......9A+8B+3C=0...........3
9*1.5B+8B+3C=0
21.5B=-3C
C=-21.5B/3
TAKING B AS 6 , WE GET A = 9...C=-43
HENCE EQN.OF REQD PLANE IS
9(X-2)+6(Y+2)-43(Z+3)=0
9X+6Y-43Z-135=0
Your answer should be in the form Ax + By + Cz - D = 0.
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