given the two planes
3x+4y-5z=12
4x+y+z=15
how can I find the symmetric equation of the line of intersection of the two planes.
i got the cross product=9i-23j-13k
but why is the point (3,2,1) a point on both the planes?
given the two planes
3x+4y-5z=12
4x+y+z=15
how can I find the symmetric equation of the line of intersection of the two planes.
i got the cross product=9i-23j-13k
but why is the point (3,2,1) a point on both the planes?
Hello, statman101!
The cross-product approach is the best way.
Here's another (primitive and longer) approach.
(Use it when you're desperate.)
And we have: .
Substitute into [2]: .
So we have: .
On the right side, replace with a parameter
. . . . .
These are the parametric equations of the line through
. . with direction vector: .
The symmetric equation is: .
Note: There are fifteen brizillion different forms of the symmetric equation.
. . . . .This is just one of them.