# intersecting lines

• September 9th 2009, 03:27 AM
tdat1979
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.
• September 9th 2009, 04:43 AM
hjortur
Quote:

Originally Posted by tdat1979
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.

a)

$v * u = |v||u|cos(\theta)$

So all you need is 2 vectors that align with those 2 lines. E.g.
$v = \left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\ri ght) , u = \left(\frac{dx}{ds},\frac{dy}{ds},\frac{dz}{ds}\ri ght)$

Hope that helps, but if you still have problems I'll try to help some more.
• September 9th 2009, 05:01 AM
HallsofIvy
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.
• September 9th 2009, 05:09 AM
Calculus26
This exact question was answered on "Question on Intersecting Lines"

and you also posted this same question again on " lines/planes?"