1. ## Lines question

If anyone could explain how the following is done without a grapher, it would be greatly appreciated:

Determine if the two lines A and B are parallel, skew, or intersecting. If they are intersecting, find the point at which they intersect.

A: x = 13 + 12t, y = -7 + 20t, z = 11 - 28t
B: x = 22 + 9s, y = 8 + 15s, z = -10 - 21s

2. Originally Posted by meebo0129
If anyone could explain how the following is done without a grapher, it would be greatly appreciated:

Determine if the two lines A and B are parallel, skew, or intersecting. If they are intersecting, find the point at which they intersect.

A: x = 13 + 12t, y = -7 + 20t, z = 11 - 28t
B: x = 22 + 9s, y = 8 + 15s, z = -10 - 21s
the direction vector to A is <12,20,-28>
the direction vector to B is <9,15,-21>

note that <12,20,-28> = (12/9)*<9,15,-21>

since the direction vectors are proportional, the lines are parallel

3. Originally Posted by meebo0129
If anyone could explain how the following is done without a grapher, it would be greatly appreciated:

Determine if the two lines A and B are parallel, skew, or intersecting. If they are intersecting, find the point at which they intersect.

A: x = 13 + 12t, y = -7 + 20t, z = 11 - 28t
B: x = 22 + 9s, y = 8 + 15s, z = -10 - 21s
These are in fact the same line. Put t=(3/4)(1+s) and substitute into the
equations for the first line, and you should find that the equations become
those for the second.

RonL

4. Hello, meebo0129!

Determine if the two lines A and B are parallel, skew, or intersecting.
If they are intersecting, find the point at which they intersect.

$A:\;\begin{array}{ccc}x &= &13 + 12t \\ y & = & \text{-}7 + 20t \\ z & = & 11 - 28t\end{array}$

$B:\;\begin{array}{ccc}x & = & 22 + 9s \\ y & = & 8 + 15s \\ z & = & \text{-}10 - 21s\end{array}$

The direction vector of $A$ is: . $\langle 12,\,20,\,\text{-}28\rangle \:=\:\langle3,\,5,\,\text{-}7\rangle$

The direction vector of $B$ is: . $\langle9,\,15,\,\text{-}21\rangle \:=\:\langle3,\,5,\,\text{-}7\rangle$

. . Therefore: . $A \parallel B$

The point $(22,\,8,\,-10)$ is on line $B$.
. . Is it on line $A$?

We have: . $\begin{Bmatrix}13 + 12t & = & 22 \\ \text{-}7 + 20t & = & 8 \\ 11 - 28t & = & \text{-}10\end{Bmatrix}\quad\Rightarrow\quad t \,=\,\frac{3}{4}$ . . . . Yes!

Therefore, lines $A$ and $B$ are coincident.