Math Is Fun Forum

Shelled

Member

Registered: 2014-04-15

Posts: 44

Lines and points

Using the equations of two lines:
(1+t, 1-3t, 2+2t) and (3s, 1-2s, 2-s).

Find the distance between the two lines and find the uniquely determined points on the two lines that are this distance apart.

I found the distance to be 1/srt(3), having trouble with finding the two points though.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,134

Re: Lines and points

hi Shelled,

How did you get the distance without finding the points first? ** I can only see how to get this by points first.

Call the lines A and B, and the line joining them C.

step 1.  Use the dot (scalar) product on the directions of A and of B to get the direction of C, perpendicular to both.

step 2.  Form an equation for points on C by equation for points on A with t as parameter add parameter r times direction of C

step 3.  Form the equation for B using parameter s.

step 4.  Find where the line C crosses B with three component equations and three unknowns r, s, t.  Solve for all three.

step 5.  The value of t will give the point on A, and s will give the point on B.

step 6. Compute the length of C.  You can do r times  'direction of C'.  (note:  I got your answer )

Bob

** This is not a criticism.  I'm just interested in your method as it must be different to mine.

Last edited by Bob (2015-06-20 03:47:15)

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Shelled

Member

Registered: 2014-04-15

Posts: 44

Re: Lines and points

Hi bob,

I think I did step 1 correctly (?) and got: -s+7ts+5+t = 0, however having trouble with following step 2.

Just with how I did it to get the distance:
I put the two lines into the form
(1,1,2)+t(1,-3,2) and (0,1,2)+s(3,-2,-1)

then used, distance = (v(dot)w) /|w|

where v = v1- v2 = (1,1,2)-(0,1,2)
and w = d1 X d2 = (1,-3,2)X(3,-2,-1)
and subbed everything in to get the answer

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Lines and points

Hi;

Using the equations of two lines:
(1+t, 1-3t, 2+2t) and (3s, 1-2s, 2-s).

Find the distance between the two lines and find the uniquely determined points on the two lines that are this distance apart.

I found the distance to be 1/srt(3), having trouble with finding the two points though.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,134

Re: Lines and points

hi bobbym,

Is it your sanity we're checking?  Don't worry.  I get that too. 

Shelled:  7ts ?  Where did that come from ?

Direction of A: (1, -3, 2)     direction of B: (3, -2, -1)     direction of C: (x, y, z)

So using the dot product twice:

x -3y +2z = 0
3x -2y -z  = 0  => 6x - 4y - 2z = 0

Add to get:

7x - 7y = 0  => x = y

You only want a vector that represents the direction so it is OK to use x = y = 1 ... and then find z.

Any point on A can be found from the equation of the line

v = (1, 1, 2) + t(1, -3, 2)

Let's say that t takes the value required for A crossing C.

Then to get to any point on C we start at (1, 1, 2), go to (1, 1, 2) + t(1, -3, 2), and then an additional r(1, 1, 1) along C

So the equation of line C is:

v = (1, 1, 2) + t(1, -3, 2) + r(1, 1, 1)

For B

v = (0, 1, 2) + s(3, -2, -1)

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob