Shortest distance between A(2,1,-2) and the line x=3-2t;y=-4+3t;z=1+2t.

[gikiian]

Homework Statement

Find the shortest distance between A(2,1,-2) and the line having parametic equations:
x=3-2t;
y=-4+3t;
z=1+2t.

Homework Equations

After derivation:
d = |AB| sin( arccos( (AB.BC)/|AB||BC| ) )

B and C are points on the line found by putting random values for t.
For t=0 -> B(3,-4,1)
For t=1 -> C(1,-1,3)

The Attempt at a Solution

Plugging the values in the above equation, we get:
d = 6 sin (arccos 11/6sqrt(17) )Using the calculator gives the correct answer, but there is another way which I can't figure out, in which you get the answer irrespective of sin and arccos.

What is it?

Thanks in advance :)

 

[Dick]

You could write down the distance squared between A and (x,y,z), then differentiate with respect to t and look for a minimum if you know calculus.

 

[HallsofIvy]

Another way to do this, and I think simpler, is to find the plane perpendicular to the given line, passing through the given point. The point on the line closest the given point- from which the shortest distance from the given point to the line is measured- is where the line intersects that plane.

The plane perpendicular to the line x=3-2t; y=-4+3t; z=1+2t, through the point (2, 1, -2) is given by -2(x- 2)+ 3(y- 1)+ 2(z+ 2)= 0. Putting the equations for the line into that we get -2(3- 2t- 2)+ 3(-4+ 3t- 1)+ 2(1+ 2t- 2)= 4t+ 9t+ 4t+ 4- 12+ 4= 25t- 4= 0. find the point corresponding to that t and the distance between that poiunt and (2, 1, -2).

 

[LCKurtz]

Another purely vector method is to make a unit vector out of the direction vector:

$$\hat D = \frac 1 {\sqrt{17}}\langle-2,3,2\rangle$$

and a vector across from any point on the line to the point off the line, for example from (3,-4,1) to (2,1,-2):

$$\vec V = \langle -1,5,-3\rangle$$

Then the distance is the magnitude of the cross product:

$$d = |\vec V \times \hat D|$$

All the sine and cosine stuff is hidden in the vector operations.