# Thread: Vectors (Parametric) Problem Urgent.

1. ## Vectors (Parametric) Problem Urgent.

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 also 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.

What would be the intercept point if the two screw lines were in the same plane. (i know they are not. but how would you find the point where the two lines make an X (top-view))

2. You're on it so far. Keep going. For notation's sake, let the two vector equations represent lines: $X(t)=(4+2t,8+3t,-1-4t)$ and $Y(s)=(7-6s,2+s,-1+2s)$ . You correctly determined that the line sought will be of the form $Z(u)=(a,b,c)+u(1,2,2)$ for some "starting vector" $(a,b,c)$. Notice that the choice for the starting vector is not unique, it can be any point on the line $Z$. Since we know $Z$ intersects $X$, without losing generality, let $Z(0)=(a,b,c)=X(t)$ for some $t$. In other words, line $Z$ "starts" at its intersection with $X$ and progresses in the direction $(1,2,2)$ towards $Y$. At some value of $u$, $Z(u)$ will intersect line $Y$. We now have the ingredients to set up a system of equations: There exists integers $t,s,u$ such that $X(t)+u(1,2,2)=Y(s)$ . This yields a 3x4 matrix with $t,s,u$ variables that can be reduced to find a unique solution. I'll let you do the grunt work for yourself, but the answer, so you can check it, is $Z(u)=(2,5,3)+u(1,2,2)$. (By the way, what would it mean if said matrix were singular? )