Subject: Calculus 3

Given two lines determine whether the lines are parallel, intersecting, or skew. If intersecting, find point of intersection.

$L1: \frac{1}{2}(x-6) = \frac{1}{3}(y-12) = \frac{1}{7}(z-4)$

$L2: \frac{1}{8}(x-1)=\frac{1}{24}(y+3)=\frac{1}{28}(z-5)$

The answer in the back says they are parallel, however when I tried to solve it on my own I reached the conclusion that they are not parallel. I compared the directional vectors:

$\frac{1/2}{1/8} = \frac{1/3}{1/24} = \frac{1/7}{1/28}$

Simplifies to

$4 = 8 = 4$

Which means they are not parallel? Did I do this correctly or did I make a mistake?

  • 1
    $\begingroup$ Could be a typo in the book. The $1/3$ was meant to be a $1/6$, or the $1/24$ was meant to be $1/12$. $\endgroup$ – Gerry Myerson Sep 11 '16 at 1:27
  • $\begingroup$ You are correct $\endgroup$ – Shailesh Sep 11 '16 at 1:29
  • $\begingroup$ You're correct. But the way you've written your post would suggest that the direction vectors would be $(\frac12,\frac13,\frac17)$ and $(\frac18,\frac1{24},\frac1{28})$. They're actually $(2,3,7)$ and $(8,24,28)$. (The logic is the same, though.) $\endgroup$ – Semiclassical Sep 11 '16 at 1:32
  • $\begingroup$ @Semiclassical so the directional vectors are the reciprocals of the coefficient? $\endgroup$ – asdfghjkl Sep 11 '16 at 1:34
  • $\begingroup$ Correct. To see this, note that you can parametrize $(x,y,z)$ by taking all three equations in $L_1$ and setting them equal to $t$. Solving, this gives $(x,y,z)=(2,3,7)t+(6,12,4)$. So one generates this line by starting at the point $(6,12,4)$ and moving forward/backward along the direction of $(2,3,7)$. $\endgroup$ – Semiclassical Sep 11 '16 at 1:58

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