Vectors - show that the lines intersect

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Hi, I'm currently revising for a maths exam and I'm stuck on the following question:

Show that the lines:

r = (i+j+k) + s(i+2j+3k)

r = (4i+6j+5k) + t(2i+3j+k)

Intersect.

My work so far:

Let (i+j+k) + s(i+2j+3k) = (4i+6j+5k) + t(2i+3j+k)

So (i) 1+s = 4+2t

(j) 1+2s = 6+3t

(k) 1+3s = 5+t

I'm unsure where to go from here, any help would be appreciated.

Regards

smn
 
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You have to find values for s and t and show that they satisfy all three vector component equations (i,j & k)

~H
 
So, you have three equations that must be satisfied for an intersection.
Is there a pair of values (s,t) that satisfies all three equations simultaneously?

[Try using, say, equation (i) with equation (j), then the result with (k) etc...] Once you determine a pair (s,t), check that it satisfies each equation.
 
Well, if they intersect, then they must by necessity have a point in common. You started on that above. If they have a point in common, what can you say about possible solutions to the system of equations you found?
 
Thanks for the prompt replys.

I realize that you have to solve for s and t and these values should equal if the lines intersect.

I was unsure what to do next with the 3 equations in order to solve for s and t.
I'm now going to try using simultaneous equations, as mentioned, to try and solve for s and t.

Regards

smn
 
smn said:
I'm now going to try using simultaneous equations, as mentioned, to try and solve for s and t.

That's the way to go!

~H
 
You have two variables s and t. You should be able solve two of the equations for them. Do those two values also satisfy the third equation?
 
Yes, i worked out that s=1 and t= -1. I then sub'd these values into the 3rd equation and it satisfied this also ( 4=4).

Thanks for all your help

Regards

Sam
 
Now, the follow-up question...
can you show that there exists an intersection WITHOUT solving explicitly for s and t?
 
Using pre-calculus methods? Only way I can think of is to show they are coplanar and not parallel.
 
daveb said:
Using pre-calculus methods? Only way I can think of is to show they are coplanar and not parallel.

Are use of the dot- and cross-product operations considered pre-calculus?
Note that the OP has already written lines in parametric vector form:
[tex]\vec A=\vec A_0 + s\vec U[/tex]
[tex]\vec B=\vec B_0 + t\vec V[/tex]
which is already somewhat advanced by introductory standards.
 
It wan't when I was in HS, but that was back in the late 70s.