Find the shortest distance between a point and a line in 3 dimensions

Question

Find the shortest distance between a point and a line in 3 dimensions:
A) Let P = (x0, y0, z0) and define the line, L, not containing P by x = x1 + a t , y = y1 + b t, z = z1 + c t, with <a,b,c> a UNIT vector.

Find a formula for the square of the distance, D, from P to any point on the line, in terms of t.

Answer 1

Let U = ⟨a,b,c⟩, Q = ⟨x₁,y₁,z₁⟩ for convenience.

 

The distance from P = ⟨x₀,y₀,z₀⟩ to any point on the line is given by the distance formula.

 

Therefore:

 

D²(t) = (at + x₁ - x₀)² + (bt + y₁ - y₀)² + (ct + z₁ - z₀)²

 

= |U|² t² + 2U·(Q - P)t + |Q - P|²

 

This is shortest at the vertex of the parabola D²(t) which is at:

 

t = -U·(Q - P)/|U|²

 

Let's plug this in to get the shortest distance.

 

D²_min =  [U·(Q - P)]²/|U|² - 2[U·(Q - P)]²/|U|²+ |Q - P|²

 

= |Q - P|² - [U·(Q - P)]²/|U|²

 

In the case of |U|=1 as in your problem, D²_min = |Q - P|² - [U·(Q - P)]²