Shortest distance between two skew lines

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SUMMARY

The shortest distance between two skew lines, L1 and L2, can be calculated using the distance formula involving the cross product of their direction vectors. For L1, defined as (x,y,z)=(4,−2,−2)+t(1,1,−3), and L2, defined by points (−2,−2,0) and (−4,−5,0), the direction vectors are (1,1,−3) and (-2,-3,0) respectively. The correct application of the distance formula requires accurate computation of the cross product and the correct points for L2, which was initially miscalculated. The final distance formula yields a result of 120/√472.

PREREQUISITES
  • Understanding of skew lines in three-dimensional geometry
  • Familiarity with vector operations, specifically cross products
  • Knowledge of the distance formula for skew lines
  • Ability to derive parametric equations from given points
NEXT STEPS
  • Study the derivation of the distance formula for skew lines in 3D geometry
  • Practice calculating cross products of vectors in three-dimensional space
  • Explore parametric equations and their applications in geometry
  • Review examples of skew lines and their properties in vector calculus
USEFUL FOR

Students studying geometry, particularly those focusing on three-dimensional vector analysis, as well as educators teaching concepts related to skew lines and distance calculations in space.

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Homework Statement


What is the shortest distance between these two lines?

L1:(x,y,z)=(4,−2,−2)+t(1,1,−3)
L2: The line through the points (−2,−2,0) and (−4,−5,0)

Homework Equations


distance formula

The Attempt at a Solution



I thought I was on the right track but apparently not.

For L1, I took 2 arbitrary t's to get 2 points on the line (which looking back, I think might be a wrong way to approach)

With t=1 and t=3, I got the points (5,-1,-5) for t=1 and (7,1,-11) for t=3.
P2-P1 = (2,2,-6) for L1

Now since the points for L2 are given:
P2-P1 = (-2,-3,0) for L2

Now I can take the cross product of the two lines:

<br /> \left| \begin{array}{ccc} i &amp; j &amp; k \\ 2 &amp; 2 &amp; -6 \\ -2 &amp; -3 &amp; 0 \end{array} \right| = -18i -12j -2k<br />Plugging this all into the distance formula equation I get:

\frac{-4(-18)-5(-12)+6(-2)}{\sqrt{-18^2-12^2-2^2}} = \frac{120}{\sqrt{472}}<br />

However, the answer is wrong. Any idea what I did wrong?

Thanks :o
 
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For one thing, if I take the cross the two direction vectors of your lines, I get (-18,12,-2). And for another thing, where did you get the (-4,-5,6) you plugged into the distance formula?
 
I totally mixed up the logic for the algorithm, but I figured it out. I got the parametric equation from the 2 points, and from there solved it. Sorry for the trouble.
 

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