Answer:
1a. L = A +t·AB = (-4, 3, 10) +t(6, 2, -18)
1b. (x, y, z) = (6t -4, 2t +3, -18t +10)
1c. (x +4)/6 = (y -3)/2 = (z -10)/-18
2. the lines are coincident
Step-by-step explanation:
You want the equation for the line through A and B written in vector, parametric, and symmetric forms. You want to know the relationship of the given lines.
1. A, B
The vector equation for the line can be written as ...
[tex]L = A +t\cdot\overrightarrow{AB}[/tex]
The vector AB has direction ...
(B -A) = (2, 5, -8) -(-4, 3, 10) = (2 +4, 5 -3, -8-10)
AB = (6, 2, -18)
Then the vector equation is ...
L = (-4, 3, 10) +t(6, 2, -18)
The expansion of this gives the parametric equation:
(x, y, z) = (-4 +6t, 3 +2t, 10 -18t)
Solving for t in each case gives the symmetric equation:
(x +4)/6 = (y -3)/2 = (z -10)/-18
2. L1, L2
For lines in vector form, the same line will be had for any linear function of the parameter (t). Here, we can replace t in the first equation by (-3/2t -3). Expanding this gives ...
L1 = (5, -4) +(-3/2t -3)(2, -4)
L1 = (5, -4) +(-3t -6, 6t +12) = (5 -6, -4+12) +t(-3, 6) = (-1, 8) +t(-3, 6)
This is exactly the equation for L2, so the lines are coincident.
