Shortest Distance from a point to a line (in 3 dimensions)

How do I find the shortest distance from a point to a line given the point and two points on the line (in 3 dimensions)?

I'm given a point P (which is (4, −1, −7 )), and a line (−2t, −1t, 3t). I know there is the formula for finding the distance and stuff but the thing screwing me up here is the line format. Any help is appreciated.

You need to determine the distance from the line to the point P for the parameter t. This is formed by taking the norm of the vector:

R(t) := r(t) - P = (-2t-4,-t+1,3t+7)

The norm (i.e. distance to the line from the point P) is

x(t) := ||R(t)|| = √( (2t+4)² + (1-t)² + (3t+7)² )

To minimize this, note that since the distance is strictly positive quantity, the minimum of x(t)² will coincide since

dx(t)²/dt = 2x(t)dx(t)/dt

Therefore, we take the derivative and set it to zero

dx²(t)/dt = 4(2t+4) - 2(1-t) + 6(3t+7) = 28t + 56 = 0

This is solved for t=-2 (notice the second derivative is positive for all t ensuring we are at a minimum) and so

min(x(t)) = √( (-4+4)² + (1+2)² + (-6+7)² ) = √10

Busy right now but hoping this helps.

Line: = t*<-2, -1, 3>

Or r(t) = t*<-2, -1, 3>

OR r(t) = <0, 0, 0> + t*<-2, -1, 3>

OR symmetric, solving for t:

x = -2t
y = -t
z = 3t

x/(-2) = t
y/(-1) = t
z/3 = t

So, x / (-2) = y / (-1) = z / 3

| a x b | = | a | | b | sin Θ....pick a point on the line { origin is a good point } and

form b = OP = < 4 , - 1 , - 7 > ; a = < -2 , - 1 , 3 >...distance is |OP| sin Θ = | a x b | / | a |