1aw sorry i knew i was gonna make a mistake. editing it.
Find eq of plane passing through (-1,1,1) and (1,-1,1) and perpendicular to plane x + 2y + 2z = 5
plz give the steps in the solution
sorry for dumb question but somehow i am getting stuck plz help.
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7 Answers
1well , we know the d.cs of the plane given it's (1,2,2) so we should find (a,b,c) such that ,
a + 2b + 2c = 0 .
eqn of any plane passing through , any point is gievn by
a(x-x1) + b ( y-y1) +c(z-z1) = 0.
substitugin for these values . we get it I think.
1A(x+1)+B(y-1)+C(z-1)=0 - (i)
put x=1,y=-1,z=1
A(2)-B(2)+C(2)=0
since the plane is perpendicular to x+2y+2z=5
therefore
A+2B+2C=0
A = B = C = λ
|-2 0| |0 2| |2 -2|
|2 2| |2 1| |1 2 |
u get
A=-4λ
B=-4λ
C=6λ
substitute in (i)
-4x-4-4y+4+6z-6=0
4x+4y-6z=-6
2x+2y-3z=-3
I suppose the method is right.
1aw !! [11]
plz check this step :-
"A(x+1)+B(y-1)+C(z-1)=0 - (i)
put x=1,y=-1,z=1
A(2)-B(2)+C(2)=0"
all u should get is A= B
1357In ankit's solution
put (1,-1,1) in (i)
we get A=B=K(say)
Also A+2B+2C=0
So C=-3K/2
so eqn is
K(x+1)+K(y-1)+(-3K/2)(z-1)=0
x+y-3z/2+3/2=0
2x+2y-3z+3=0
1thats what i got in the end only my method was a bit messed up [2]
Thanks for this convincing solution [1]