Math Is Fun Forum

djbip

Member

Registered: 2008-03-27

Posts: 13

Vector problem

Can someone please help me with the following problems.

1. "Find paramtric equations of the line that passes through the point of intersection of L1 and L2, and that is perpendicular to both."

L1: (x-4)/2=(y -8 )/3=(z+1)/-4
L2: (x-16)/-6=(y-2)/1=(z+1)/2

i know that the result for this one is a pretty ugly one...

2. "Find parametric equations of a line that intersects both L1 and L2 at right angles"

L1: [x,y,z]=[4,8,-1] + t[2,3,-4]
L2: (x-7)/-6=(y-2)/1=(z+1)/2

I need the results in max 8 hrs....after that it is useles!! I hope someone will be able to help me

Thank You!

Last edited by djbip (2008-03-27 19:46:35)

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Vector problem

This post is fun to read but it does have some mistakes
according to Wolfram web site on "cross product".
See post after this one for Wolfram way.
I've been thinking instead of looking up a vector product or something.
And I think a way might be possible to do some rock, paper, scissors here with
the x,y, and z-axes.     x goes to y goes to z goes to x with some conversions and negations
like in 2-D.  In 2-D, you negate the reciprocal of slopes.
Here, maybe dx/dy goes to -dz/dy, where y goes to z and x goes to y ?
Then dy/dz could become -dx/dz, where z goes to x and y goes to z ?
And perhaps dz/dx could become -dy/dx where x goes to y and z goes to x ?
It could be circular?

5 minutes later theory:
Compare components of two lines at same time.
This is all hypothetical so far (wrong?)
Lines are marked 1, 2, 3;  3 being the perpendicular answer line.
dx1 perpend dy2 -> dz3
dy1 perpendicular dz2 -> dx3
dz1 perpendicular dx2 -> dy3

Using above negative reciprocals for perpendicularity, we
are hoping this is right:

formula#RPS (rockpaperscissors)
-dy2 / dx1 -> dz3
-dz2 / dy1 -> dx3
-dx2 / dz1 -> dy3

These "d" values are like components.

Now let's give a go (try) this problem with new funny formula#RPS:
#1)
L1: (x-4)/2=(y -8 )/3=(z+1)/-4
L2: (x-16)/-6=(y-2)/1=(z+1)/2
components are in the denominators in this format.
We will ignore the numerator offsets and intersection at this time.
dx1 = 2;   dy1 = 3  ;  dz1 = -4
dx2 = -6  ;  dy2 = 1 ; dz2 = 2
Plug it into RPS:
dz3 = -1/2  ;   dx3 = -2/3   ;  dy3 = 6/(-4)  = -3/2
These (if right?) can go in denominator or parametric equations:
t = (x - a)/dx3
t = (y - b)/dy3
t = (z - c)/dz3
But I gotta go shovel snow and I'm not sure how to
do the simultaneous equations to get the intersection
(a,b,c) right now.  Bye!
And this, if right, might be hopefully like matrix workings?

Last edited by John E. Franklin (2008-03-28 06:18:14)

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: Vector problem

The mistake I made was to assume that the 2-D
negative reciprocals was going to hold true for
3-D, and apparently it doesn't, according to
Wolframs equation, which seems different because
it uses 4 components instead of only 2 components
to determine each "d" value or component along an axis.

(My dx is like "x hat" in Wolfram "cross product")
(dx3,dy3,dz3 is the direction of the perpendicular answer we are looking for)

Wolfram way:

dx3 = dy1 dz2 - dy2 dz1

dy3 = dz1 dx2 - dz2 dx1

dz3 = dx1 dy2 - dx2 dy1

Now let's try #1) problem the right way, hopefully!!
#1)
L1: (x-4)/2= (y-8 )/3= (z+1)/-4
L2: (x-16)/-6= (y-2)/1= (z+1)/2
components are in the denominators in this format.
We will ignore the numerator offsets and intersection at this time.
dx1 = 2;   dy1 = 3  ;  dz1 = -4
dx2 = -6  ;  dy2 = 1 ; dz2 = 2
Plug into Cross Product way:
dx3 = (3)(2) - (1)(-4)      (for this one, I think the first term of the answer with y&z rotates around to x, just go around yzx by following xyzxyzxyz, but starting on y)
dy3 = (-4)(-6) - (2)(2)   (your solving for dy, so think the first term is z&x, like xyzxyz and start on z)
dz3 = (2)(1) - (-6)(3)   (solving for dz is easiest because it is just xyz, you don't go around from z back to x)

dx3 = 10 ;  dy3 = 20  ;  dz3 = 20

parametric equations going thru point (a,b,c):
t = (x - a)/dx3
t = (y - b)/dy3
t = (z - c)/dz3

We still need to find point (a,b,c), that is, if the two
lines do intersect.  If they don't intersect, it is cool
and awesome that you can still have a "perpendicular"
line.  (Try making lines in the air with your 2 arms and see!)

L1: (x-4)/2= (y-8 )/3= (z+1)/-4
L2: (x-16)/-6= (y-2)/1= (z+1)/2

I reason that you can't just set the right sides of the L1 and L2 equations together.
You have to keep L1 all in sync if you multiply and subtract and stuff.
Because even though z = -1 for both equations, doesn't have
any relevance that I can see, after plotting in the air with my arms.
So in keeping the 3 sides of the equations together,
here we go:
But the (z+1) can get cancelled out with multiplying and subtraction,
as long as we do the same thing to the x and y parts of the equations.

L1 + (1/2)(L2):    (adjusting denominators with numerators as needed)
(5x - 8)/12 = (5y - 22)/6 = 0

Notice that the 3 sides of the equations L1 = -1.2 and 3 sides of L2 = 2.4  when the lines intersect,
not that that is important or anything.

x = 8/5  ; y = 22/5  ;  z = 3.8

So then, the perpendicular line's equation can be:

L3:

t = (x - (8/5))/10
t = (y - (22/5))/20
t = (z - 3.8)/20

or stated as perhaps:

L3: [x,y,z]=[8/5,22/5,3.8] + t[10,20,20]

(Unfortunately, I haven't checked the work, and the 8 hour requirement has past)
(I started pointing out the lines in the air at my desk, and it does
make okay sense within a radius of about 5 units from expected intersection.)
The angle of the 3rd line also seems to be correct within 45 degrees just using
my hands to rough it out in the air.
I double checked the cross product and it comes up either -10, -20, -20, or
+10, +20, +20, so either one is fine since the line goes both ways forever.

Last edited by John E. Franklin (2008-03-28 07:55:39)

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igloo myrtilles fourmis