How can I find the distance between two parallel lines that are in vector form?

Question

I have a question to do, and I keep getting the answer wrong.

I need to know the distance between these parallel lines:

L1: [x,y,z] = [1, 3,-4] + s[2,-5,2]

L2: [x,y,z] = [4, 0, 2] + t[2,-5,2]

The answer is 4.58, but I have no idea how to get that.

Thank you to everyone in advance.

Answer 1

The formula for distance between parallel lines in $3D$ is $$\frac{|\vec b ×(\vec a_{2}-\vec a_{1})|}{|\vec b|}$$ where $L_1=\vec a_1+\lambda \vec b,L2=\vec a_2+\lambda \vec b$ can you continue from here.

Answer 2

Hint:

If $A=(1,3,-4)$ and $B=(4,0,2)$, $\vec p=(2,-5,2)$ and $\theta$ is the angle between $\vec AB$ and $\vec p$, then the required distance is $|\vec{AB}\sin\theta|$.

Use cross product.

Answer 3

If you can't remember a fancy distance formula, you could solve the old-fashioned way:

minimize the distance between $(4,0,2)+t(2,-5,2)=(4+2t,-5t,2+2t)$ and $(1,3,-4).$

The distance squared is $(3+2t)^2+(3+5t)^2+(6+2t)^2=33t^2+66t+54,$

which has its minimum when $66t=-66$, i.e., $t=-1$,

so the distance squared is $33(-1)^2+66(-1)+54=21.$

Answer 4

The distance between the lines is measured perpendicularly to them. So, pick a point on each line and project the segment joining them onto a plane perpendicular to them. The distance between the lines is equal to the length of the segment’s projection.