SOLUTION: Find the equation of the plane which contains the line: (x,y,z) = (0,2,-3) + t(-3,3,0) and is parallel to the line: (x,y,z) = (-1,-3,0) + t (-1,-2,3) Write your answer

Question 62455: Find the equation of the plane which contains the line:
(x,y,z) = (0,2,-3) + t(-3,3,0)
and is parallel to the line:
(x,y,z) = (-1,-3,0) + t (-1,-2,3)
Write your answer in the form A x + -1 y + C z = D, and give the values of A, C and D as your answer.

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
Find the equation of the plane which contains the line:
SAY L1.. (x,y,z) = (0,2,-3) + t(-3,3,0)
and is parallel to the line:
SAY L2... (x,y,z) = (-1,-3,0) + t (-1,-2,3)
Write your answer in the form A x + -1 y + C z = D, and give the values of A, C and D as your answer.
LET THE DRS OF THE REQUIRED PLANE P BE L,M,N.
IT IS PARALLEL TO L2 AND CONTAINS L1.HENCE NORMAL TO PLANE IS PERPENDICULAR
TO THE 2 LINES . HENCE
-3L+3M+N*0=0......3L=3M.....L=M............1
-L-2M+3N=0...........................2
-3M+3N=0.....M=N.
LET L,M,N BE 1,1,1
SO EQN. OF PLANE IS X+Y+Z=C
SINCE THE PLANE CONTAINES L1,(0,2,-3) IS A POINT ON THE PLANE
SO 0+2-3=C=-1
HENCE EQN. OF PLANE IS X+Y+Z+1=0

You may edit the question. Maybe convert formulae to the same formula notation as in
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