Find the image of the pointP(3, 5, 7) in the plane 2x+y+z= 6.

Solution

The feet of perpendicular of the point (3, 5, 7) on the given plane is the point in which the line through (3, 5, 7) perpendicular to the given plane meets it.

Now the direction ratios of the normal to the given plane are 2, 1, 1.

∴ Equation of the line through P(3, 5, 7) perpendicular to the given plane are

Any point on this line is (3 + 2r, 5 + 4, 7 + r)

If it lies on the given plane 2x + y + z = 6, then we have

2(3 + 2r) + (5 + r) + (7 + r) = 6  or 6r = −12  or r = −2

∴ feet of perpendicular is (3 + 2r, 5 + r, 7 + r), where r = −2

i.e., (3 − 4, 4, 5 − 2, 7 − 2)

i.e., (−1, 3, 5)

If points of image are (α, β, γ) then

∴ α = −5, β = 1, γ = 3

∴ required image is (−5, 1, 3).