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## 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 :)