Question

How to find canonical form of equation of a line then general form is given

Answer 1

General form:
\[3x+2y-4z+1=0\]
\[5x-y+2z-4=0\]

Answer 2

@amistre64 perhaps you could help me with this problem since you helped me before? :)

Answer 3

you are given 2 plane equation right? and they want the line of intersection?

Answer 4

what is " canonical form of equation of a line"

Answer 5

http://www.scribd.com/doc/72014999/18/The-Canonical-Equations-of-a-Straight-Line-The-Parametric

this looks helpful :)

Answer 6

"Since the given straight line is determined by the intersection of the planes"

cross the normal to determine the direction vector of the line of intersection, then all it amounts to is finding a point that is common to both planes to attach it to

Answer 7

x 3 5 x: 2(2)-(-4)(-1) = 0
y 2 -1 -y: 3(2) - 5(-4) = -26
z -4 2 z: 3(-1) - 5(2) = -13

and since vectors are scalable, lets reduce that to <0,2,1>

now to find a point that is common to both planes; lets zero out one of the dimension leaving us with an intersection of 2 lines

lets try to find the common point in the xy plane (no z)

3x+2y+1=0
5x - y - 4=0

3x+2y+1=0
10x -2y -8=0
------------
13x - 7 = 0; x = 7/13


5(7/13) - (13y/13) - 4(13/13)=0

35 - 13y - 52 = 0 ; y= -17/13

so we have a point (7/13, -17/13, 0 ) that is on the line

Answer 8

and of course a little late idea after all this is just to define 2 points, one from different planes and define the vector between them instead of the cross product of the normals

dime or dozen, either way :)

Answer 9

Thanks for help, @amistre64 . But you stopped at the right at the end were I am having the problem.
How can I write the cannonical form if one coordinate of the vector is 0? It can't look like this, can it?
\[\frac{x- \frac{ 7 }{ 13 } }{ 0 } = \frac{y+ \frac{ 7 }{ 13 } }{ 2 } = \frac{z }{ 1 }\]

Answer 10

i believe the conical forms are:

x = 7/13 + 0t
y = -17/13 + 2t
z = 0 + 1t

Answer 11

x really aint a function of t, its just a constant

Answer 12

That should be parametric form (or something similar, don't know what's the name in english)

Answer 13

heres what i see on a 3d plotter

http://web.monroecc.edu/manila/webfiles/calcNSF/JavaCode/CalcPlot3D.htm

Answer 14

how about we just make x=7/13 in the 2 generals and then combine them

Answer 15

21/13+2y-4z+1=0

35/13-y+2z-4=0

this is a line in the yz plane

Answer 16

since x becomes "undefined" it might have to be a side note to the yz parts

Answer 17

Well, I still need cannonical form. I doubt there is any way to get rid of that zero in the vector coordinates, so what should I do?

Answer 18

Perhaps I don't need to find the canonical form. I think knowing the direction vector is enough to solve the real problem

Answer 19

http://www.netcomuk.co.uk/~jenolive/vect17.html

towards the end of the page is a similar issue; and they just leave it as a side not "z=0"

Answer 20

I see. Thanks, it'll be useful :)

Answer 21

its like tryng to say; put x=3 into slope intercept form ...

Answer 22

spose we try that in conan form:

x = 3 + 0t
y = t

x-3 y
--- = ---
0 1

yet we know that the line x=3 exists so i would say that since y is not a function of x, and x is not a function of t, we resort to the notation

y
--- and x=3 to express a line where such that (3,y) are all the points
1