Question

Problem Set 3 1E: 7

Answer 1

I looks like problem set 3 solutions is missing the answer to 1E:7.
For the first part, is the answer:
\[abs\left( \frac{ v_{1} X v_{2} }{ \left|v_{1} X v_{2} \right| } . \left( p_{2} - p_{1}\right) \right)\]
???

Where v1 is the direction of line 1 and v2 is the direction of v2, and p1 and p2 are points on line1 and line2 respectively?

Thanks!
-Gene

Answer 2

They hint says that "the shortest line segment joining the two skew lines will be perpendicular to both of them". So v1 X v2 will be perpendicular to both lines (since "skew" lines are nonparallel this is legal, otherwise if they were parallel then the cross product would make a zero vector).

I then normalize v1 x v2 and project a connecting vector onto the normalized ortho vector. The absolute value of this ought to be the shortest distance between the two lines, right?

Thanks!
-Gene

Answer 3

Yes, that is what I found. For the distance between 2 lines, each on adjacent faces of a unit cube, I found a distance of
\[ \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \]

Answer 4

Ah yeah, I got the same answer for part (b).

Thanks so much Phi!