How far is the point (7,4,1) from the line that passes through the points (6,4,1) and (6,3,-1)?

To begin with, we should find the equation of the line. To do this we appoint a position vector and a direction vector. The position vector can simply be one of the two points we are give, i.e. (6,4,1). To find the direction vector, we subtract one of our points from the other: (6,4,1)-(6,3,-1)=(6-6,4-3,1+1)=(0,1,2). Thus our line is r=(6,4,1)+t(0,1,2), where t is a numeric parmeter.

Now, the shortest difference between a point and a line is always when the line s joining the point and the line r is perpendicular to r. This means that R.S=0, as cos90=0, and R and S are the direction vectors of the lines r and s respectively.

We know that the direction vector S can be derived by subtracting the given point (7,4,1) from an arbitrary point on the line r, which we are trying to determine, i.e. S=(6+0t, 4+t, 1+2t) - (7,4,1) = (-1,t, 2t).

So, R.S=(0,1,2).(-1,t, 2t) = 0x1 + 1xt + 2x2t = t+4t = 5t 

We know that R.S=0=5t, so it follows that t=0, so the point on the line r which is closest to (7,4,1) is when t=0 for line r, so (6,4,1).

To find the distance between two points, we use pythagoras:

sqrt[(7-1)^2 + (4-4)^2 + (1-1)^2]=sqrt[36]=6.

So the point (7,4,1) is 6 units away from the line passing through the points (6,4,1) and (6,3,-1).