Question
Find the image of the point (1, 2, 3) in the plane x + 2y + 4z = 38.
Solution
The equation of plane is
x + 2 y + 4 z = 38 ...(1)
Direction ratios of the normal to the plane are 1, 2, 4
Let M be foot of perpendicular from P(l, 2, 3) to the plane.
Now PM is a straight line which passes through P( 1, 2, 3) and has direction ratios as 1,2,4.
∴ its equations are
Any point M on line is (r + 1, 2 r + 2, 4 r + 3)
∵ M lies on plane (1)
∴ (r + 1) + 2 (2 r + 2) + 4 (4 r + 3) = 38
∴ r + 1 +4r + 4 + 16r + 12 = 38
∴ 21 R = 21 ⇒ r = 1
∴ M is (1 + 1,2 + 2,4 + 3) i.e. (2,4,7)
Let N(α, β, γ) be image of P in the plane (1) so that M is mid-point of PN.
∴ image is (3, 6, 11).
