# vector calc...point of intersection

• Sep 17th 2009, 12:52 PM
holly123
vector calc...point of intersection
show that the two lines x=(-1/2 t, 2/3 t +1, t- 3/2) and x= (t- 3/2, 2t+3, 3t+ 9/2) intersect. find an equation for the plane that they determine.

i have no clue how to do this i keep getting stuck.. the professor's hint was:
Part (1): show that the given lines intersect. In order to do this, you should rename one of the variables given in the expressions for these lines. The book writes both lines using the same variable, t. In order to find out where these two lines meet, you need to first rename the variable in the second line, say, to s. Then set the two expressions equal which sets the three vector components equal and then solve the system of equations that results. You will find a value for t and a value for s. Plugging these values in to their respective lines produces (or should produce, if your math is correct) the same point -- the point of intersection.

Part (2) then asks you to find the equation of a plane containing these intersecting lines, which I think you can do without a hint

anyone want to start me off? i changed the variables in the second equation to s's, but what do i set equal? the entire equations or just the x's, y's, and z's? i'm so lost (Crying)
• Sep 17th 2009, 01:17 PM
Calculus26
Before going into the details you may want to see my web site
about intersecting and skew lines in 3 space --notes and animations.

Vector Valued Functions

Now with your problem --you can think of the lines as the trajectories of particles. They may either collide in which case they reach the pt of int at the same time. The paths could cross but the particles reach the pt of int at different times whichj is why you need 2 variables.

(-1/2 t, 2/3 t +1, t- 3/2)

(s- 3/2, 2s+3, 3s+ 9/2)

equating the x coord -1/2t = s- 3/2

t= -2s+3

equate the y cooords

2/3t +1 = 2s +3

2/3(-2s+3)+1 = 2s + 3

-4s + 9 = 6s +9

s= 0

so on L2
(s- 3/2, 2s+3, 3s+ 9/2) -> (-3/2,3,9/2)

since t= -2s+3 it follows when s = 0 t= 3

On L1
(-1/2 t, 2/3 t +1, t- 3/2) -> (-3/2,3,9/2) as it should
• Sep 17th 2009, 01:28 PM
Calculus26
Elaboration
By the way if s = t the particles collide

and if for the given values of s and t the z coords don't match you woud have skew lines
• Sep 17th 2009, 03:08 PM
holly123
thanks i was actually on the right track that time, but when i got s=0 for some reason i thought i was doing it wrong. how do i do the second part? is the point (-3/2, 3, 9/2) and the normal is (-3/2, -4/3, -2)?
not too sure about the normal..i took the difference of the t coordinates in the given problem
• Sep 17th 2009, 03:16 PM
holly123
or do i take the cross product to find the normal?
• Sep 17th 2009, 03:25 PM
Calculus26
you can use any point on either line (-3/2, 3, 9/2) is fine

to find N take the crosspro duct of the 2 direction vectors

-1/2 i +2/3 j + k and i + 2j +3k
• Sep 17th 2009, 04:04 PM
holly123
(0, 5/2, -5/3) ?
• Sep 17th 2009, 04:23 PM
Calculus26
yes

you can also multiply by 6/5 to eliminate fractions and simplify and use

(0,3,-2)