How To Determine If Vector Intersects A Plane Defined as 'z = 0'
Your calculations are right, it does indeed intersect. Try drawing a picture to help your intuition! In this case, the vector emanating from $p_2$ moves up and down only in the z-direction, so must intersect the plane $z=0$ at some point
This is guaranteed because $v2$ is the normal vector, if you want to guarantee that the line doesn't intersect the plane, try using a parallel vector
ImJustACowLol
Bachelor's degree in Medical Informatics. Current occupation: iOS developer for ING Group (largest bank in The Netherlands) Primary programming languages: Swift, Objective-C, Java Also worked with: Angular 6+, JavaScript, R, Python, Typescript, Ruby, SQL
Updated on April 29, 2020Comments
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ImJustACowLol over 3 years
So I'm rather new to vector/3d mathematics so i'm probably missing something but...
If we have a point p at location (0, 0, 0) (the origin) from which a vector v departs, which is defined as <0, 0, 1> (so the vector only travels in the z-direction).
To get the plane of which v is the normal-vector I could use the following formula:
- a1(x - x0) + a2(y - y0) + a3(z - z0) = 0
if we fill it in we get:
- 0(x - 0) + 0(y - 0) + 1(z - 0) = 0
- (z-0) = 0
- z = 0
This plane should be a plane that spans the entirity of the x and y axis and does not move into the z-direction at all. Now, if we have the point p2 = (1, 2, 3) with vector v2 = <0, 0, 1> then we know that v2 from p2 should not intersect the plane that we have made. Yet, when I parametrize v2, fill the z in in the plane-formula then I get a formula that is solvable;
V2 parametrized is:
- <1 + 0t, 2 + 0t, 3 + 1t>
If we substitute x, y and z in our plane-formula then we get...
- 3 + 1t = 0
- t = -3
And if I feed this back into the parametrized vector then the point of intersection is:
- (1 + 0*-3, 2 + 0*-3, 3 + 1*-3) =
- (1, 2, 0)
The problem here is, there should be no point of intersection at all... What am I doing wrong?
Sorry if I made any noob-errors.
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Siong Thye Goh about 6 yearsmath.meta.stackexchange.com/q/5020/306553 mathjax reference for typing maths.
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ImJustACowLol about 6 yearsBut then I don't understand this: the end-point of v2 from p2 is the coordinate (1, 2, 4). The line that can be drawn between (1, 2, 3) and (1, 2, 4) does not intersect the plane that I defined. How do I mathematically figgure out that the line between those points do not intersect with the plane?
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TheMathsGeek about 6 yearsAh, I see what your problem is. In your head, you have just got the $p_2 + v_2 = (1,2,4)$. When you parametrise the line you end up with a whole set of points given by $(1,2, 3+t)$ where $t$ can vary. Setting $t=-3$ in this shows the point on the plane $z=0$
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ImJustACowLol about 6 yearsSo I can assume that when t < 0, that the point of intersection is actually "behind" the vector or point from which the vector eminates?
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ImJustACowLol about 6 yearsIn the answer by amd he said that t is a negative number in the case of v2. Does this mean that whenever a t-value is negative (and thus the point of intersection is opposite to the vector) that the plane is behind point and vector for which the t-value has been calculated?
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TheMathsGeek about 6 yearsYeh! Start from the point $p$. You can then choose to move either way along the vector $(0,0,t)$. If $t >0$ in this case you move 'away' from the plane. However, if you choose to go with $t < 0$ you get 'closer' to the plane until you hit it at $t= -3$ (and then keep going past it for t=-4, -5, ...$
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ImJustACowLol about 6 yearsdamn, I'm actually getting the hang of this... Thanks man! You just made my day :) I'm building a 3D-engine just for fun, and obviously I need quite some vector/3d calculus in order to do so. I've had some of the theory at the uni, but most of the knowledge faded away. Fortunately people like you are around to help developing such mathematical skills :) Cheers bud!
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TheMathsGeek about 6 yearsMore than welcome! Wolfram has a great parametric curve plotter for simple curves here: wolframalpha.com/widgets/… until you get the hang of it
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amd about 6 years@ImJustACowLol Exactly. Think of the vector as a velocity and $t$ as time. If $t<0$ the intersection was in the past and you’re moving even farther away from the plane.