# Intersection of a plane 3D Geo

• Apr 26th 2011, 04:09 PM
aoibha
Intersection of a plane 3D Geo
Could anybody pease explain to me where I have gone wrong in the sum below the answer is ment to be t = 0.5

The line p(t) starting at [2,-1,1] and going in the direction [1,3,-2] and the plane defined by 3x+y -2z = -2

using the formula: t =
-n.s + n.po
_________
n.v

I took s = [2,-1,1] v = [1,3,-2] n = [3,1 -2] and n.p0 = -2

-n.s = [3,-1,2].[2,-1,1] = [-6,1,2]

n.v = [3,1,-2].[1,3,-2] = [3,3,4]

[-6,1,2] -2
_________
[3,3,4]

[-8,-1,0]
_______
[3,3,4]

t = [-2.6, -0.3,]
t = [ -2. 9]
• Apr 26th 2011, 06:48 PM
Soroban
Hello, aoibha!

Sorry, I don't understand that formula you used.

You didn't tell us what we're looking for.
Is it the intersection of the line and the plane?

Quote:

The line p(t) starting at [2,-1,1] and going in the direction [1,3,-2]
and the plane defined by: .3x + y - 2z .= .-2

. . . . . . . . . . . . . . . . . . . . . . . x .= . 2 + t
The equations of the line are: . y .= .-1 + 3t
. . . . . . . . . . . . . . . . . . . . . . . z .= . 1 - 2t

Substitute into the equation of the plane:

. . 3(2 + t) + (-1 + 3t) - 2(1 - 2t) .= .-2

. . . . . . . .6 + 3t - 1 + 3t - 2 + 4t .= .-2

. . . . . . . . . . . . . . . . . . . . . . 10t .= .-5

. . . . . . . . . . . . . . . . . . . . . . . . t .= .-0.5

The intersection is: .(1.5, -2.5, 2)

• Apr 27th 2011, 08:10 AM
aoibha
Thanks! I was just trying to find t.
• Apr 29th 2011, 11:38 PM
earboth
Quote:

Originally Posted by aoibha
Could anybody pease explain to me where I have gone wrong in the sum below the answer is ment to be t = 0.5

The line p(t) starting at [2,-1,1] and going in the direction [1,3,-2] and the plane defined by 3x+y -2z = -2

using the formula: t =
-n.s + n.po
_________
n.v

I took s = [2,-1,1] v = [1,3,-2] n = [3,1 -2] and n.p0 = -2

-n.s = [3,-1,2].[2,-1,1] = [-6,1,2] <--- NO! The result is -3

n.v = [3,1,-2].[1,3,-2] = [3,3,4] <--- NO! The result is 10

[-6,1,2] -2
_________
[3,3,4]

[-8,-1,0]
_______
[3,3,4]

t = [-2.6, -0.3,]
t = [ -2. 9]

Your formula works fine but ...

1. The equation of the plane is p: $\vec n \cdot \vec r = -2$

The equation of the line is $\vec r = \vec s + t \cdot \vec v$

Therefore the Point of intersection is determined by:

$\vec n \cdot (\vec s + t \cdot \vec v) = -2~\implies~t=\dfrac{-2-\vec n \cdot \vec s}{\vec n \cdot \vec v}$

2. And now comes the point where you have made a (I don't know an appropriate adjective) mistake: The result of a scalar product of 2 vectors is a real number. So using your values you'll get:

$t = \dfrac{-2-3}{10} = -\dfrac12$