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?
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.
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.
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?
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).