vectors and planes

[paulxzt]

Determine whether the lines L1 and L2 are parallel, skew, or intersecting. If they intersect, find the point of intersection.

L1 = (x-1)/2 = (y-3)/2 = (z-2)/-1
L2 = (x-2)/1 = (y-6)/-1 = (z+2)/3

i see that it is given in a symmetric equation form..

can someone help me with this ? thanks

 

[soroban]

Hello, paulxzt!

Determine whether the lines \(\displaystyle L_1\) and \(\displaystyle L_2\) are parallel, skew, or intersecting.
If they intersect, find the point of intersection.

\(\displaystyle L_1:\;\frac{x-1}{2} \:= \:\frac{y-3}{2} \:=\:\frac{z-2}{-1}\)

\(\displaystyle L_2:\;\frac{x-2}{1}\: = \:\frac{y-6}{-1}\:=\:\frac{z+2}{3}\)

Rewrite in parametric form.

\(\displaystyle L_1\!:\begin{Bmatrix}x & = & 2t+1 \\ y &=&2t+3 \\ z &=&\text{-}t+2 \end{Bmatrix}\quad L_2\!:\begin{Bmatrix}x &=& u+2 \\ y &=&\text{-}u+6 \\z &=&3u+2\end{Bmatrix}\)


\(\displaystyle \text{Their direction vectors are unequal: }\;\langle2,2,-1\rangle \:\neq\:\langle 1,-1,3\rangle\)
. . The lines are not parallel.


\(\displaystyle \text{If the lines intersect, the system: }\;\begin{array}{cccc}2t + 1 &=& u+2 & [1] \\ 2t+3 &=& -u+6 & [2] \\ -t+2 &=& 3u+2 & [3]\end{array} \;\text{ will have a solution.}\)

\(\displaystyle \text{Add [1] and [2]: }\;4t+4\:=\:8\quad\Rightarrow\quad4t \:=\:4\quad\Rightarrow\quad t \:=\:1\)
\(\displaystyle \text{Substitute into [1]: }\;2(1) + 1 \:=\:u+2\quad\Rightarrow\quad u \:=\:1\)

But these roots do not satisfy [3] . . . The lines do not intersect; they are skew.