Q)
(chatquestion)...EXAMPLE 18
Find the equation of the plane passing through the points
(-1, 1, 1) and (1,-1, 1) and ls perpendicular to the
plane x + 2y +2z 5
O Solution The equation of a plane passing through the
point (-1, 1, 1) is
a(x+1) + b(y-1) + c(z-1) = 0
...(1)
[where a, b, c are constants]
Now, the plane (1) passes through the point (1, -1, 1)
a(1 + 1) + b(-1 - 1) + c(1-1) = 0
2a - 2b = 0
or,
a - b+0 c= 0
or,
...(2)
Since the plane (1) is perpendicular to the plane
x + 2y + 2z = 5 (given), so, normals to the planes (1) and
x + 2y + 2z = 5 are are perpendicular to each other.
a + 2b + 2c = 0
..*(3)
Hence, from (2) and (3) [by cross-multiplication] we get,
9.
-k (+0, say)
%3D
3.
a = -2k, b =-2k, c= 3k in (1) we get,
- 2k(x + 1) - 2k(y-1)+3k(z-1) = 0
- 2x- 2-2y + 2 + 3z -3 = 0 [: k#0]
or,
2x + 2y-3z+3 = 0
Therefore, the equation of the required plane is
or,
2x + 2y-3z +3 = 0
...
