How to Solve a Parametric Vector Problem Involving Perpendicular Vectors?

[gomess]

Homework Statement

Homework Equations

(x,y,z)=(x₀,y₀,z₀) + t(m₁,m₂,m₃)

The Attempt at a Solution

So at first I thought that since vector P₁P₂ is at right angles to both lines, both lines must be parallel. Quickly dismissed this idea since their direction vectors are not multiples of one another. Then I thought P₁P₂ could be the cross product of both direction vectors... crossed both vectors and got P₁P₂=(-1,3,1). Not sure if that's the right approach, and not sure what to do from here. Any help would be great!

 

[SammyS]

Homework Statement

Attached thumbnail

Homework Equations

(x,y,z)=(x₀,y₀,z₀) + t(m₁,m₂,m₃)

The Attempt at a Solution

So at first I thought that since vector P₁P₂ is at right angles to both lines, both lines must be parallel. Quickly dismissed this idea since their direction vectors are not multiples of one another. Then I thought P₁P₂ could be the cross product of both direction vectors... crossed both vectors and got P₁P₂=(-1,3,1). Not sure if that's the right approach, and not sure what to do from here. Any help would be great!

Posting the image will make it more likely that your question will get attention:

I haven't worked through the problem, but it seems to me that the dot product (scalar product) may work better.

 

[gomess]

I'll try using the scalar product and see where it gets me.
Edit: No progress. I figured that since P₁ lies on L₁, some value of 't' would take me to P₁ and from there, vector P₁P₂ would take me to P₂ (assuming P₁P₂ is the cross product of the direction vectors).

 

[SammyS]

I'll try using the scalar product and see where it gets me.
Edit: No progress. I figured that since P₁ lies on L₁, some value of 't' would take me to P₁ and from there, vector P₁P₂ would take me to P₂ (assuming P₁P₂ is the cross product of the direction vectors).

Oh! Yes, that should work for the direction of $\displaystyle\vec{P_1P_2} \ .\$ You were correct about the result of <-1, 3, 1>