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.
To find the point of intersection, solve the two equation x= 3t- 3= s- 6 and y= -2= -3s-5 for s and t, then check to be sure they also satisfy z= 6t+ 7= 2s+ 1. (In three dimensions most pairs of lines don't intersect but since they ask for the point of intersection, I presume these do.)
To find the plane they both lie in, take the cross product of the "direction vectors" (that you also used in finding the angle between them) to find a vector perpendicular to both and use that as the "normal vector" to the plane.