So the distance from 𝑃=(3,−3,3) P = ( 3 , − 3 , 3 ) to the line through the points 𝐴=(−1,−4,0) A = ( − 1 , − 4 , 0 ) and 𝐵=(−3,0,−1) B = ( − 3 , 0 , − 1 ) is

Question

The distance  of a point  to the line through points  A and B is the length of the component AP of  that is orthogonal to AB, as indicated in the diagram.

So the distance from 𝑃=(3,−3,3) P = ( 3 , − 3 , 3 ) to the line through the points 𝐴=(−1,−4,0) A = ( − 1 , − 4 , 0 ) and 𝐵=(−3,0,−1) B = ( − 3 , 0 , − 1 ) is

Answer 1

The parametric equations of the line passing through the points A and B are

x = -1 -2t

y = -4 + 4t

z = -t

Let us take a point Q on the line passing through A and B. For t = -1 we get Q ( 1, -8 ,1 )

Then vector QP = < 2, 5 , 2 > and || QP || = √ 2² + 5² + 2² =√29

The length of the projection of the vector QP on the vector AB is ( 2/3) √21

Therefore the distance d of the point P to the line AB using The Pythagorean Theorem is

d² = ( √29 )² − ( ( 2/3) √21 )²

d = √(59/3)