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Equation of a line intersepting two planes

Posted 11 years ago
I know equations of two planes and I want to find the line of interseption of those. But I do not know any point on that intercepting line.
What I leart so far is that, by taking the cross product of the perpendicular vectors of the two planes, I can find the vector of the line. But without knowing a point on the line I can't find the line equation. 
Appreciate your help to solve this problem.
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The general method is as follows: Each plane (2D plane in 3D euclidean space) has a normal vector. The intersection of the two planes is a line and this line is normal to both of the normal vectors by definition. So it is the cross product of the two normals.
POSTED BY: Udo Krause
@ChinthakaRatnaweera are you asking a purely math question? This is the site dedicated to posts ONLY about Wolfram Technologies. We do not address general math questions.
POSTED BY: Vitaliy Kaurov
Aculally I am looking for a general way of solving this.
How to find the range of values that satisfy t (parameter)?
This link explains a general method.
But I don't really follow Gellert et al's general approach.
Great someone can explain this more.
No need for a point. Use parameteric representation
 ClearAll[x, y, z, t];
 p1 = x + 5 y + 2 z + 3 == 0; (*equation of first plane*)
 p2 = 2 x + 3 y + 6 z + 4 == 0; (*equation of second plane*)
 eq1 = Simplify[Inner[Subtract, Thread[2 *p1, Equal], p2, Equal]]; (*eliminate x*)
 zt = t; (*let z=t*)
 yt = y /. First@Solve[eq1 /. z -> zt, y];  (*find y in terms of t*)
 xt = x /. First@Solve[p1 /. {z -> t, y -> yt}, x]; (*find x in terms of t*)
Plot3D[z /. First@Solve[p1, z], {x, -2, 2}, {y, -3, 3},PlotStyle -> Green, Mesh -> None],
Plot3D[z /. First@Solve[p2, z], {x, -2, 2}, {y, -3, 3},PlotStyle -> LightBlue, Mesh -> None],
ParametricPlot3D[{xt, yt, zt}, {t, -3, 3}, PlotStyle -> {Thickness[.03], Red}], Axes -> None, Boxed -> False

POSTED BY: Nasser M. Abbasi
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