Skew Lines Conceptual Problem

  1. 1. The problem statement, all variables and given/known data
    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.


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

    3. 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?
     

    Attached Files:

  2. jcsd
  3. 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.
     
  4. How?
     
  5. 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.
     
  6. 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?
     
  7. 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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