# Thread: distance

1. ## distance

find the shortest distance between the point (0,0,1) and the line

(x-1)/2 = (y-2) /4 = (z-3) /4

2. Originally Posted by alexandrabel90
find the shortest distance between the point (0,0,1) and the line

(x-1)/2 = (y-2) /4 = (z-3) /4
What work have you done so far?

3. i know there is a point (1,2,3) on the line. let that be point Q and P= (0,0,1)

then PQ=(1,1,2)

distance of PQ is sq root 18.

and then im not sure what i can do alr..

4. Originally Posted by alexandrabel90
i know there is a point (1,2,3) on the line. let that be point Q and P= (0,0,1)

then PQ=(1,1,2)

distance of PQ is sq root 18.

and then im not sure what i can do alr..
Hmm, for the distance between P and Q, I get

$D=\sqrt{1^2+2^2+2^2}=3$

But this only gives an upper bound (the final answer can't possibly be greater than 3). Let the line be L. To find the shortest distance, you need to find the line perpendicular to L passing through P.

5. You can't just choose a point on the line. The point on the line closest to (0, 0, 1) is at the foot of the line perpendicular to the given line through (0, 0, 1). Of course, the perpendicular line will lie on a plane perpendicular to the given line.

The given line, (x-1)/2 = (y-2) /4 = (z-3) /4 can also be written in parametric equations by letting the parameter, t, be equal to each of those: t= (x-1)/2 so x= 1+ 2t. t= (y-2)/4 so y= 2+ 4t. t= (z-3)/4 so z= 3+ 4t.
Since the given line has "direction vector" <2, 4, 4>, any plane perpendicular to the line has equation 2x+ 4y+ 4z= C for some constant C. When x= 0, y= 0, z= 1, 2(0)+ 4(0)+ 4(1)= 4 so the equation of the plane perpendicular to the given line and containing (0, 0, 1) is 2x+ 4y+ 4z= 4.

Find the point where the given line crosses that plane. The distance from (0, 0, 1) to the point you just found is the distance from the point to the line.