Given the lines : L1: x = -1 - 2t, y = -6t, z = 8t
L2: x = 3 + t, y = 3t, z = 5 - 4t
(a) Show that the lines are parallel.
(b) Find the distance between the two lines.
(c) Find the equation of the plane containing the two lines.
(a) $\displaystyle \vec{n}_1=\left[\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\right]=[-2, -6, 8] \ \ \
\vec{n}_2=[ 1, 3, -4] \ \ \vec{n}=\frac{[1, 3, -4]}{\sqrt{26}} $
(b) Pick a point on L1, say (-1, 0, 0) and another point on L2, say (3, 0, 5). The vector is $\displaystyle \vec{r}=[4, 0, 5] \ \ \ d=|\vec{r} \times \vec{n}|$
(c) The normal vector of the plane is $\displaystyle \vec{r} \times \vec{n}$
a.
For L1 v1 = -2 i - 6j +8k
L2 v2 = i + 3 j - 4k
since v1 = -2v2 they are parallel
b. consider the vector v from (3,0,5) on L2 to (-1,0,0) on L1
Then the distance between the 2 lines is the magnitude of v - projection of v on v2
See attachment
v = 4 i + 5k
projv on v2 = (4 i + 5k)*(i + 3j -4k)/26 *(i +3j -4k)
= (-16/26)(i +3j -4k)
v - projv on v2 = (4+16/26)i +48/26j + (5-64/26)k
|v- projv on v1 | = 5.58
c. Use 2 points on L1 (-1,0,0) and (1,6,-8)
one point on L2 (3,0,5)
Let u be the vector from (-1,0,0) to (3,0,5) and v be the vector
from (1,6,-8) to (3,0,5)
Then the normal to the desired plane is N = u x v = ai +bj+ck
using any of the 3 points use a(x-x0) +b(y-y0) +c(z-z0) = 0
to generate the eqn of the plane