Thread: Calc III

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.

Originally Posted by tdat1979

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)

(b) Pick a point on L1, say (-1, 0, 0) and another point on L2, say (3, 0, 5). The vector is

(c) The normal vector of the plane is

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