Thread: Question on intersecting lines

Given the intersecting lines:

L1 : x = 3t − 3, y = −2t, z = 6t + 7
L2 : x = s − 6, y = −3s − 5, z = 2s + 1

(a) Find the (acute) angle, rounded to the nearest degree, between the lines.
(b) Find the point of intersection of the two lines.
(c) Find the equation of the plane containing the two lines.

Thanks in advance!

a For L1 v1 = 3i - 2j + 6k

L2 v2 = i - 3j + 2k

the angle q between the lines is the angle between v1 and v2:

cos(q) = v1*v2/(|v1||v2|)

b

set the x coordintates equal 3t - 3 = s -6
s = 3t+3

set the y components equal -2t = -3s -5 = -9t - 9 +5

At the pt of intersection t = -2 and s =-3 yielding (-9,4,-5) as the pt of int

c. The normal to the plane is v1 x v2 = N = ai + bj + ck. choose a pt on either line , say (-3,0,7)

Use a(x+3) +by + c(z-7) = 0 to obtain eqn of plane