- #1

Peeter

- 305

- 3

Given the parametric representation of two planes, through points P and Q respectively

[tex]

x = P + \alpha u + \beta v

[/tex]

[tex]

y = Q + a w + b z

[/tex]

Or, alternately, with [tex]u \wedge v = A[/tex], and [tex]w \wedge z = B[/tex]

[tex]

x \wedge A = P \wedge A

[/tex]

[tex]

y \wedge B = Q \wedge B

[/tex]

It's easy enough to find the intersection of the two planes by wedging, x = y, with A or B. For example, if I wedge with B I get a parametric equation for the line of intersection, provided it exists:

[tex]

x = P + \frac{(Q-P)\wedge B}{v \wedge B}}v + \alpha\frac{1}{v \wedge B}((v \wedge B) u - (u \wedge B)v)

[/tex]

Intuitively, I'd expect that one could reduce this to an expression in terms of only P, Q, A, and B without requiring an explicit pair of direction vectors in the plane (the points and the bivectors specify the planes and their position, so I'd expect the intersection to be expressable that way too).

I don't see how to obviously do the reduction I'd imagine possible here (I'm self teaching myself geometric algebra as it appears to be the natural language for much of physics which I'd like to understand better ... to get a feel for things I was going through some of the sorts of calculations learned with traditional means such as row reduction, and this was one I tried.)

[tex]

x = P + \alpha u + \beta v

[/tex]

[tex]

y = Q + a w + b z

[/tex]

Or, alternately, with [tex]u \wedge v = A[/tex], and [tex]w \wedge z = B[/tex]

[tex]

x \wedge A = P \wedge A

[/tex]

[tex]

y \wedge B = Q \wedge B

[/tex]

It's easy enough to find the intersection of the two planes by wedging, x = y, with A or B. For example, if I wedge with B I get a parametric equation for the line of intersection, provided it exists:

[tex]

x = P + \frac{(Q-P)\wedge B}{v \wedge B}}v + \alpha\frac{1}{v \wedge B}((v \wedge B) u - (u \wedge B)v)

[/tex]

Intuitively, I'd expect that one could reduce this to an expression in terms of only P, Q, A, and B without requiring an explicit pair of direction vectors in the plane (the points and the bivectors specify the planes and their position, so I'd expect the intersection to be expressable that way too).

I don't see how to obviously do the reduction I'd imagine possible here (I'm self teaching myself geometric algebra as it appears to be the natural language for much of physics which I'd like to understand better ... to get a feel for things I was going through some of the sorts of calculations learned with traditional means such as row reduction, and this was one I tried.)

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