Question

Find the image A' of the point A(1, 6, 3) in the line `x/1 = (y - 1)/2 = (z - 2)/3`. Also, find the equation of the line joining A and A'.

Answer 1

Given the equation of a line.

`x/1 = (y - 1)/2 = (z - 2)/3 = λ`

Direction ratios of the line AP are 

λ – 1, 2λ + 1 – 6, 3λ + 2 – 3

i.e., (λ – 1, 2λ – 5, 3λ – 1)

Direction ratio of line L is (1, 2, 3)

AP ⊥ L

∴ a₁a₂ + b₁b₂ + c₁c₂ = 0

1(λ – 1) + 2(2λ – 5) + 3(3λ – 1) = 0

λ – 1 + 4λ – 10 + 9λ – 3 = 0

14λ = 14

λ = 1

∴ P(1, 2(1) + 1, 3(1) + 2)

Hence coordinates of P(1, 3, 5).

Let A' is (x, y, z) then

P is the midpoint of AA'

(1, 3, 5) = `((1 + x)/2, (6 + y)/2, (3 + z)/2)`

1 = `(1 + x)/2, 3 = (6 + y)/2, 5 = (3 + z)/2`

2 = 1 + x, 6 = 6 + y, 10 = 3 + z

x = 1, y = 0, z = 7

Image A' (1, 0, 7)

Equation of the line joining A and A'

`(x - x_1)/(x_2 - x_1) = (y - y_1)/(y_2 - y_1) = (z - z_1)/(z_2 - z_1)`

`(x - 1)/(1 - 1) = (y - 6)/(0 - 6) = (z - 3)/(7 - 3)`

`(x - 1)/(0) = (y - 6)/(-6) = (z - 3)/(4)`