Question

My question is with respect to PS3-1E7. The solution appears to be missing entirely. Does anyone see something I'm missing or know where the solution might be found. Thanks.

Answer 1

Here is the problem and where the solution ought to be.

Answer 2

The asterisk means there is no solution available.

Answer 3

But a solution for 1E7 is as follows.

Formulate a general method for finding the distance between two skew lines.

The formula for a line in space i.e. 3D, is
\[ P= P_0 + t \vec{V} \]
where P0 is a point on the line, V is a "direction vector" and t is a scalar that lets us move along the line.

Given two lines
\[ P= P_1 + t \vec{V_1} \\ P= P_2 + t \vec{V_2}\]
the shortest distance between the two lines is on a line perpendicular to both directions vectors, namely
\[ \vec{N}= \vec{V_1}\times \vec{V_2}\]
and make this vector unit length:
\[ \hat{N}= \frac{\vec{V_1}\times \vec{V_2}}{|\vec{V_1}\times \vec{V_2}|}\]
next, create a vector from line 1 to line 2 (by using a point in each line): \( P_2-P_1\)

Answer 4

the distance d is
\[ d = |(P_2-P_1) \cdot \hat{N} |\]
we take the absolute value to guarantee positive values