- #1

- 63

- 0

Thank You

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter applestrudle
- Start date

- #1

- 63

- 0

Thank You

- #2

Simon Bridge

Science Advisor

Homework Helper

- 17,857

- 1,655

There is a standard proof of this written down someplace but I forget it.

You start out by finding the closest point Q on a line L1 to a specified point P.

You should notice that:

(I) Then line-segment QP is perpendicular to L1. Do you see why this must always be the case?

(II) If Q were not the closest point on L1 to P, then QP would not be perpendicular to L1.

Now consider that P is on another line L2 ...

1. If L2 is parallel to L1, then |PQ| is the distance of closest approach and PQ is perpendicular to both.

2. If L2 intersects L1, then |PQ| is not the distance of closest approach unless |PQ|=0.

3. If L2 is skew to L1 ... then we already have QP perpendicular to L1 beacuse Q is the closest point on L1 to P. But P is not the closest point to Q on L2*unless* PQ is perpendicular to L2 ...

because of statement (II) above.

i.e.

For |PQ| to be the distance of closest approach for the two skew lines, P must be the closest point on L2 to Q*and* Q must be the closest point on L1 to P ... so PQ must be perpendicular to both lines.

Since L1 and L2 are*lines* there can be only one (P,Q) pair such that |PQ| is the distance of closest approach - i.e. the solution is unique.

You start out by finding the closest point Q on a line L1 to a specified point P.

You should notice that:

(I) Then line-segment QP is perpendicular to L1. Do you see why this must always be the case?

(II) If Q were not the closest point on L1 to P, then QP would not be perpendicular to L1.

Now consider that P is on another line L2 ...

1. If L2 is parallel to L1, then |PQ| is the distance of closest approach and PQ is perpendicular to both.

2. If L2 intersects L1, then |PQ| is not the distance of closest approach unless |PQ|=0.

3. If L2 is skew to L1 ... then we already have QP perpendicular to L1 beacuse Q is the closest point on L1 to P. But P is not the closest point to Q on L2

because of statement (II) above.

i.e.

For |PQ| to be the distance of closest approach for the two skew lines, P must be the closest point on L2 to Q

Since L1 and L2 are

Last edited:

Share: