# Thread: lines in space

1. ## lines in space

For the life of me; this whole topic makes no sense at all !

The parametric vector form of the line L1 is given as r1 = u1 + rv1 ( is a real number)
where 1 u is the position vector of 1 P1 = (1,1,−3) and v1 = P1P2where P2 = (3,3,−2) .

The parametric vector form of the line L2 is given as r2 = u2 + sv2 (s is a real number)
where u2 is the position vector of P3 = (−2,0,2) and v2 = −j− k .

(a) Give the parametric scalar equations of the lines L1 and L2

(b) Find the unit vector ˆn with negative i component which is perpendicular to both L1and L2

(c) The shortest distance between two lines is the length of a vector that connects the two lines and is perpendicular to both lines. For L1 and L2
this is expressed in the vector equation r2 −r1 = tnˆ where (t is a real number) is a
parameter. Write this equation as 3 scalar equations and hence obtain a system of three linear equations for the three parameters r, s and t .

(d) Solve this system of equations for r, s and t and hence find the shortest distance between the two lines L2 and L2

(e) Find the point Q on line L1 which is closest to line L2

Thanks for any help !

2. ## Re: lines in space

Originally Posted by bluelavae

The parametric vector form of the line L1 is given as r1 = u1 + rv1 ( is a real number)
where 1 u is the position vector of 1 P1 = (1,1,−3) and v1 = P1P2where P2 = (3,3,−2) .
I completely lost with your notation. Here is what I taught for over 30yrs in the US.
Vector form: $<1,2,4>+t<-2,5,3>$
parametric form: $\left\{ {\begin{array}{*{20}{l}} {x = 1 - 2t} \\ {y = 2 + 5t} \\ {z = 4 + 3t} \end{array}} \right.$
symmetric form: $\dfrac{x-1}{-2}=\dfrac{y-2}{5}=\dfrac{z-4}{3}$.

That is three form of the same line through point $(1,2,4)$ with direction $<-2,5,3>$.

Can you say what your question means?

3. ## Re: lines in space

Ah sorry, hopefully this image shows

4. ## Re: lines in space

Originally Posted by bluelavae
Ah sorry, hopefully this image shows
Apparently the mathematics community in your part of the world use a different set of terms.
I do not understand it.

But if $L_1:P+tD~\&~L_2:Q+sE$ are two non-parallel lines then the distance between them is:
$\dfrac{{\left| {\overrightarrow {PQ} \cdot \left( {D \times E} \right)} \right|}}{{\left\| {\left( {D \times E} \right)} \right\|}}$