Question

A(1, 0, 4), B(0, -11, 13), C(2, -3, 1) are three points and D is the foot of the perpendicular from A to BC. Find the co-ordinates of D.

Answer 1

Equation of the line passing through the points (x,₁, y₁, z₁) and (x₂, y₂, z₂) is 
`(x - x_1)/(x_2 - x_1) = (y - y_1)/(y_2 - y_1) = (z - z_1)/(z_2 - z_1)`
∴ the equation of the line BC passing through the points B (0, –11, 13) and C)2, –3,1) is
`(x - 0)/(2 - 0) = (y + 11)/(-3 + 11) = (z - 13)/(1 - 13)`

i.e. `x/(2) = (y + 11)/(8) = (z - 13)/(-12) = lambda`     ..(Say)
AD is the perpendicular from the point  A(1, 0, 4) to the line BC.
The coordinates of any point on the line BC are given by 
x = 2λ, y = –11 + 8λ, z = 13 – 12λ
Let the coordinates of D be (2λ, – 11 + 8λ, 13 – 12λ) ...(1)
∴ the direcion ratio of AD are
2λ – 1, λ 11 + 8λ – 0, 13 – 12λ –  4
i.e. 2λ – 1, – 11 + 8λ, 9 – 12λ
The direction ratios of the line BC are 2, 8, – 12.
Since AD is perpendicular to BC, we get
2(2λ – 1) + 8(– 11 + 8λ) – 12(9 – 12λ) = 0
∴ 4λ – 2 – 88 + 64λ – 108 + 144λ = 0
∴ 212λ – 198 = 0
∴ λ = `(198)/(212) = (99)/(106)`
Putting λ = `(99)/(106)` in (1), the coordinates of D are 
`(198/106, -11 + 792/106, 13 - 1188/106)`

i.e. `(198/106, (-374)/106, 190/106)`,

i.e. `(99/53, (-187)/53, 95/53)`.