how to find the distance of a point from a given line in 3d - Maths - Three Dimensional Geometry

Question

how to find the distance of a point from a given line in 3d

Anuradha Sharma · Accepted Answer

Here is the example for the same Find the coordinates of the foot of the perpendicular from the point (2,3,-8) to the line (4-x)/2 = y/6 = (1-z)/3 Also find the perpendicular distance from the given point to the given line

Let L be the foot of the perpendicular drawn from the point P(2, 3, – 8) to the given line The coordinates of a general point on $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}$ are given by, $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} = λ$ x = 4 – 2λ, y = 6λ, z = 1 – 3λ Let the coordinates of L be (4 – 2λ, 6λ, 1 – 3λ) ∴ Direction ratios of PL are proportional to 4 – 2λ – 2, 6λ – 3, 1 – 3λ – (– 8) i e , 2 – 2λ, 6λ – 3, 9 = 3λ Direction ratios of the given line are proportional to 2, 6, 3 Now, PL is perpendicular to the given line $∴ 2 (2 - 2 λ) + 6 (6 λ - 3) + 3 (9 - 3 λ) = 0 \Rightarrow 4 - 4 λ + 36 λ - 18 + 27 - 9 λ = 0 \Rightarrow 23 λ + 13 = 0 \Rightarrow 23 λ = - 13 \Rightarrow λ = \frac{- 13}{23}$ So, the coordinates of L are $4 - 2 (\frac{- 13}{23}), 6 (\frac{- 13}{23}), 1 - 3 (\frac{- 13}{23}) = \frac{118}{23}, \frac{- 78}{23}, \frac{62}{23}$ ∴ Length of the perpendicular, $\sqrt{(\frac{118}{23} - 2)^{2} + (\frac{- 78}{23} - 3)^{2} + (\frac{62}{23} + 8)^{2}} = 11 33$