Intersecting Planes: Is It Possible?

[Philosophaie]

I have two 3D planes: A1*x + B1*y + C1*z + D1 = 0 and A2*x + B2*y + C2*z + D2 = 0.

If you set them equal to each other it should be at the intersection.

This leads to another Plane: (A1 - A2)*x + (B1 - B2)*y + (C1 - C2)*z + (D1-D2) = 0.

What I want is the line of intersection in vector and parametric form.

This is not a homework problem. It is for personal knowledge.

Is this Possible?

 

[Ssnow]

Yes, you can consider the vector perpendicular to your plane $(A_{1}-A_{2},B_{1}-B_{2},C_{1}-C_{2})$ and the vector perpendicular to the other plane that you want the intersection line, for example the first $(A_{1},B_{1},C_{1})$. Doing the exterior product $(A_{1}-A_{2},B_{1}-B_{2},C_{1}-C_{2})\wedge (A_{1},B_{1},C_{1})$ you will find the vector of the parametric equation of the intersection line. Now you need only a common point in order to find the parametric equation of the line.

Ssnow

 

[fresh_42]

I have two 3D planes: A1*x + B1*y + C1*z + D1 = 0 and A2*x + B2*y + C2*z + D2 = 0.

If you set them equal to each other it should be at the intersection.

This leads to another Plane: (A1 - A2)*x + (B1 - B2)*y + (C1 - C2)*z + (D1-D2) = 0.

What I want is the line of intersection in vector and parametric form.

This is not a homework problem. It is for personal knowledge.

Is this Possible?

Yes, but the outcome can be empty, a line or the entire plane if your input equations happen to describe the same plane. You also cannot simply set them equal the way you did, because that means you lose an entire condition: they both have to be true, simultaneously. This means the use of one equation as a substitute in the other has to eliminate one of the parameters, coordinates.

 

[jbriggs444]

I have two 3D planes: A1*x + B1*y + C1*z + D1 = 0 and A2*x + B2*y + C2*z + D2 = 0.

If you set them equal to each other it should be at the intersection.

If you set the left hand sides equal to one another you change the solution set. Instead of getting all points which satisfy equation 1 and all points which satisfy equation 2, you get all points which are equidistant (in some appropriately weighted sense) from plane 1 and plane 2.

 

[Philosophaie]

$(A_{1}-A_{2},B_{1}-B_{2},C_{1}-C_{2})\wedge (A_{1},B_{1},C_{1})$
$=((A_{1}-A_{2})*B_{1}-(B_{1}-B_{2})*A_{1})*(\hat x \wedge \hat y) + ((C_{1}-C_{2})*A_{1}-(A_{1}-A_{2})*C_{1})*(\hat z \wedge \hat x) + ((B_{1}-B_{2})*C_{1}-(C_{1}-C_{2})*B_{1})*(\hat y \wedge \hat z)$

Is this the correct exterior product?

 

[Ssnow]

... Is this the correct exterior product?

yes now you must continue $\hat{x}\wedge \hat{y}=\hat{z}$ and so on ...
Ssnow