Skew Lines Conceptual Problem

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  • #1
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Homework Statement


Suppose L1 and L2 are skew lines. Is it possible for a non-zero vector to be perpendicular to both L1 and L2? Give reasons for your answers.


Homework Equations


I know that skew lines are not parallel or intersect. Also, they don't lie on the same plane.

The Attempt at a Solution


I say that it is possible.
Picture a line on the x axis on (x,0,0) and a line on the y-axis exactly above the previous line but with height or z=5.
I can easily draw a vector that is perpendicular to both L1 and L2.

Am I correct?
 

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Answers and Replies

  • #2
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yes you are correct
suppose you have the equations of two skew lines, you can always deduce the equation of the line passing through and perpendicular to them.
 
  • #3
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How?
 
  • #4
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Given [tex]
\begin{array}{l}
(l_1 ):r = \left( {\begin{array}{*{20}c}
{a_1 } \\
{a_2 } \\
{a_3 } \\
\end{array}} \right) + t_1 \left( {\begin{array}{*{20}c}
{b_1 } \\
{b_2 } \\
{b_3 } \\
\end{array}} \right) \\
(l_2 ):r = \left( {\begin{array}{*{20}c}
{a^' _1 } \\
{a^' _2 } \\
{a^' _3 } \\
\end{array}} \right) + t_2 \left( {\begin{array}{*{20}c}
{b^' _1 } \\
{b^' _2 } \\
{b^' _3 } \\
\end{array}} \right) \\
\end{array}[/tex]
Let A, B be 2 arbitrary points lying on (l1) and (l2) respectively, then write the position vector of them.
Write the equation of [tex]
\overrightarrow {AB}[/tex]
Since AB is perpendicular to (l1) and (l2): [tex]
\overrightarrow {AB} .\left( {\begin{array}{*{20}c}
{b_1 } \\
{b_2 } \\
{b_3 } \\
\end{array}} \right) = 0
[/tex]
and [tex]
\overrightarrow {AB} .\left( {\begin{array}{*{20}c}
{b^' _1 } \\
{b^' _2 } \\
{b^' _3 } \\
\end{array}} \right) = 0
[/tex]
Solving the equations simultaneously, you'll find t1 and t2, which are later be used to compute the coordinates of A and B. The vector [tex]
\overrightarrow {AB}[/tex] is what you're after.
 
  • #5
307
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What reasons can I give to my answer?
I know that it can be done, because it's geometrically possible to picture. But, how do I put it into academic words that can be considered a correct answer?

Also, I know how to find a point A and B in L1 and L2. It's just giving an arbitrary value to their parameters. In addition, I know how to find the segment uniting points AB perpendicular to both lines.
But, how do I solve the equations simultaneously?
Do you mean:
a1+(b1)t=a1'(b1')s
a2+(b2)t=a2'(b2')s
a3+(b3)t=a3'(b3')s

and then find "t" and "s"? How do I later find the coordinates of AB?
 
  • #6
132
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It's just giving an arbitrary value to their parameters.
It's not. You are supposed to write down the position vectors of them in terms of t1 and t2. Then you'll have 2 equations of 2 unknowns (t1 and t2), which is solvable. Finding the coordinates is just a matter of substitution now.

For example, given (l): r=(1 2 3) + t(3 2 1). For every point M lying on (l), its coordinate is of the form (1+3t, 2+2t, 3+t).
 

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