# Find the vector, parametric, symmetric equations of a line

1. Mar 9, 2013

### NATURE.M

1. The problem statement, all variables and given/known data

Find the vector, parametric and symmetric equations of a line that intersect both line 1 and line 2 at 90°.

L1:
x = 4 + 2t
y = 8 + 3t
z = -1 − 4t

L2:
x = 7 - 6t
y = 2+ t
z = -1 + 2t

2. Relevant equations

vector, parametric, symmetric equations of line in R3 and cross product equation.

3. The attempt at a solution

I obtained the direction vector for the line (L3) that intersects L1 and L2. It is [1,2,2].

And I let the point of intersection between L3 and L1 be:
[x1,y1,z1]=[4,8-1]+t[2,3,-4], tεℝ

And let the point of intersection between L3 and L2 be:
[x2,y2,z2] = [7,2,-1]+s[-6,1,2], sεℝ

And now to find the scalar multiple of direction vector of the L3 that intersects point 1 and point 2.
So, [x1,y1,z1]+n[1,2,2]=[x2, y2, z2], nεℝ

And when I solve I obtain s=-1, n=-1,..
But then I saw a different approach in which the individual used,

[x1,y1,z1]-[x2, y2, z2]=n[1,2,2], nεℝ

And this yields, s=1, n=1, t=-1.

So my question is which method is accurate/correct, or does it not really matter a great deal?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 9, 2013

### SammyS

Staff Emeritus
Take your equation,  [x1,y1,z1] + n[1,2,2] = [x2, y2, z2]  and rearrange it to

[STRIKE][x1,y1,z1] - n[x2, y2, z2] = -n[1,2,2] .[/STRIKE]

[x1,y1,z1] - [x2, y2, z2] = -n[1,2,2] .

This is equivalent to the other individual's result !

Edited post to fix typo.

Last edited: Mar 9, 2013
3. Mar 9, 2013

### NATURE.M

But there not equivalent. If you rearrange, [x1,y1,z1]+n[1,2,2]=[x2, y2, z2]
We obtain:[x1,y1,z1]-[x2, y2, z2]=-n[1,2,2] or [x2, y2, z2]-[x1,y1,z1]=n[1,2,2], which are not equivalent to [x1,y1,z1]-[x2, y2, z2]=n[1,2,2]---subtraction is not commutative.
And plus, if they were equivalent how would the s,t and n values obtained vary??
So once again does it not matter which method you use, or is there something I'm missing?

4. Mar 9, 2013

### SammyS

Staff Emeritus
Your n is equal to -n of your 'friend' , but that just makes the equations exactly equivalent, if you have the same t as he/she.

5. Mar 9, 2013

### NATURE.M

Okay I see what your saying. So then it really doesn't matter which equation I use.
Thanks.

6. Mar 9, 2013

### NATURE.M

One other question, when you substitute the s=-1 value into the equation
[x2,y2,z2] = [7,2,-1]+s[-6,1,2] the answer varies when you input the other s value.
But would it still be right??