Distance from a point to a line?

Let u = (4, 5, 1) - (0, 5, 1) = (4, 0, 0) and a = (0, 3, 1). Then the distance between the point (4, 5, 1) and the given line is

|| u - projection of u onto a || = || u - u.a/||a||^2*a || = || u || = 4.

Unfortunately, one of the other answers was voted down when most of the work is, in fact, correct. Zach used a more general approach that could be applied in many other situations, especially when a list of formulae is not handy. Anyways, the problem is in the calculation of the cross product which should be BA x BC = (0, -4, 12). We now have || BA x BC ||/||BC|| = 4√(10)/√(10) = 4.

Call the given point A. So A(4, 5, 1)

Find two points on the line. By letting t = 0 we get B(0, 5, 1) and t = 1 we get C(0, 8, 2)

Get the vector BA = <4, 0, 0> and the vector BC = <0, 3, 1>

The two vectors make an angle. If we drop an orthogonal line from the point to the line we get a right triangle. The length of this line is sin(p)|BA|

Where |BA| is the magnitude of BA and p is the angle between BA and BC.

sin(p) = |BA x BC| / |BA||BC|

So sin(p)|BA| = |BA x BC| / |BC|

BA x BC = <0, -4, 12>

So, |BA x BC| = 4sqrt(10)

|BC| = sqrt(10)

Thus sin(p)|BA| = 4 which is the distance from the point to the line.

Edit: Fixed miscalculation. ><