Originally Posted by **chinthaka** I have two equations of planes . Lets say ax + by + cz = k and dx +ey + fz = k'
When solving these two equations I should get an equation for a line. Now if I assume z = t then I can find x, y in terms of t parameter. That will give me a parametric equation for the line. But the problem is when a = b = 0 or ae = bd denominator become zero. So this solution is not correct for general case. I am not physics/math expert. I don't understand what to do next. Thank you if any one can guide me to find the equation of intersection between two plains.
Thanks
Chin |

hello!

**CASE #1:**
a = b = 0

cz = k

so z = k/c >>>> equation of a plane parallel to the xyplane

if we substitute this in the equation of the 2nd plane,

dx + ey + fk/c = k'

ey = -dx - fk/c + k'

y = (-d/e)x + (-fk/ec + k'/e)

for your 2nd case, i don't know >.< i'm still figuring it out..

EDIT:

okay. i think i have it figured out..

**CASE #2:**
i'm not sure if you've already tackled this in class.. but

a normal vector of plane 1 is <a, b, c>

a normal vector of plane 2 is <d, e, f>

if the 2 planes are parallel to each other, they have no line of intersection. To know if this is the case, the normal vectors of the 2 planes should have the same direction.. thus, they are scalar multiples of each other:

<a,b,c> = k<d,e,f>

Hence, k = a/d = b/e = c/f

rearranging this part:

a/d = b/e

we get:

ae = bd

so the 2 planes have no intersection