The Student Room Group
Reply 1
i) assuming that the xy plane is horizontal...that means that the normal is the z direction...

0x + 0y +1z = d

e.g. z=8 or z = -8

ii) if the plane is vertical then its normal must be perpendicular to the z direction.

if the normal is ai + bj + ck and the z direction is 0i + 0j + 1k then it follows that c =0 in order for the scalar product equal zero.

so the normal is ai + bj + 0k

and the cartesian equation is ax + by = d
Reply 2
thanks for that.

but how can u find the cartesian equations of the line of intersection of the plane: 2x - y + 2z = 4 and a general horizontal plane: cz + d = 0?

is it: x = 0, y = 0, z = c ?
Reply 3
davemarkey
thanks for that.

but how can u find the cartesian equations of the line of intersection of the plane: 2x - y + 2z = 4 and a general horizontal plane: cz + d = 0?

is it: x = 0, y = 0, z = c ?

Nope,
Actually, you can write the equation of the line as the simultaneous of 2 equations, which are 2 equations of 2 planes.

Or
You can find 2 different points P and Q which lie in both planes.
e.g
Choose P, let x = 0, you will have y = 2z - 4 (from the 1st plane), and z = -d/c(from 2nd plane)
Choose Q, let y = 0 .... you'll get the cartesian co-ordinate of Q.
Then find PQ vector, and write it like
r = OP + t.PQ where t is parameter, O(0,0,0).

Or
It can be written like
(x-xP)/(xQ-xP) = (y-yP)/(yQ-yP) = (z-zP)/(zQ-zP)
Reply 4
thanks BCHL85

i tried that and i got the cartesian equations, but in term of c.

for example: (3xc)/(4c+2d) = (yc)/(2d+4c) + 1 and z=-d/c

please could u tell me whether that looks right
thanks a lot BCHL85

I could also do with some help on davemarkey's question :p:

Is the answer he got right???
Reply 6
davemarkey
thanks for that.

but how can u find the cartesian equations of the line of intersection of the plane: 2x - y + 2z = 4 and a general horizontal plane: cz + d = 0?

is it: x = 0, y = 0, z = c ?


rearranging the second plane equation we find that z = -d/c

putting this into the first plane equation we find that

2x - y = 4 + 2d/c

or y = 2x -(4 + 2d/c)

or (y - 0 )/1 = ( x - ( 2 + d/c ))/0.5 and z = -d/c