VECTORS QUESTION (parametric equation)

[037810]

For example

[x,y,z]=[4,8,-1]+t[2,3,-4]

[x,y,z]=[7,2,-1]+s[-6,1,2]


I did...

x=4+2t=6s
y=8+3t=2+s
z=-1-4t=-1+2s

and got t = -6/5 and s=12/5

so since no interception and not parallel, it's a skew line.

But if they were in the same plane.. how would you find the intersection?

This is confusing me.

thank you.

 

[rock.freak667]

You'd find it in the same manner. Just that from two equations you get values of t and s that would satisfy the third equation.

 

[037810]

yah but these two lines are already skew? so the third equation does not satisfy.

if the screw lines of the diff planes come down to make just one plane, they'd obviously have an intersection right?

how would you find that intersection?

 

[037810]

because I'm trying to solve this question:

find parametric eq of a linethat intersects both lines, [x,y,z]=[4,8,-1]+t[2,3,-4]

[x,y,z]=[7,2,-1]+s[-6,1,2] at 90 degrees.

So with direction vectors, i cross multiplied to get [1,2,2].

I just need to know the point that the line goes through to complete the equation.

*I'm kinda sketchy though the intersection of the two skewlines will be where the line will pass through..

I'm not sure how to solve this problem..

 

[rock.freak667]

Bearing in mind that I have not vectors in a long time. I would guess that you should, do something like find <(4+2t),(8+3t),(-1-4t)>.<1,2,2>=0 and solve for t. Then put that back into the line for t to get the point.

 

[037810]

t's would cancel out in that case..