Find the perpendicular distance of the point (1, 0, 0) from the line (x - 1)/2 = (y + 1)/-3 = (z + 10)/8

Question

Find the perpendicular distance of the point (1, 0, 0) from the line \(\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8}\) Also find the co-ordinates of the foot of the perpendicular.

(x - 1)/2 = (y + 1)/-3 = (z + 10)/8

Answer 1

Let PM be the perpendicular drawn from the point (1, 0, 0) to the line \(\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8}\) = λ …(Say)

The coordinates of any point on the line are given by x = -1 + 2λ, y = 3 + 2λ, z = 8 – λ 

Let the coordinates of M be 

(-1 + 2λ, 3 + 3λ, -1 – λ) …..(1)

The direction ratios of PM are 

-1 + 2λ – 2, 3 + 3λ + 3, -1 – λ – 1 

i.e. 2λ – 3, 3λ = 6, -λ – 2

The direction ratios of the given line are 2, 3, 8. 

Since PM is perpendicular to the given line, we get

2(2λ – 3) + 3(3λ + 6) – 1(-λ – 2) = 0 

∴ 4λ – 6 + 9λ + 18 + λ + 2 = 0 

∴ 14λ + 14 = 0 

∴ λ = -1 

Put λ in (1), the coordinates of M are 

(-1 – 2, 3 – 3, -1 + 1) i.e. (-3, 0, 0). 

∴ length of perpendicular from P to the given line 

= PM

Alternative Method : 

We know that the perpendicular distance from the point

Substituting teste values in (1), we get 

length of perpendicular from P to given line 

= PM