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?

- September 21st 2010, 11:45 AMstatman101Symmetric equation of the line of intersection of the two 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? - September 21st 2010, 11:49 AMUnknown008
You have to get any point which is found on both planes. The point you take can be any point, as long as it satisfies both equations.

I could take for example,

If I take x = 1, I get z = 17/3 and y = 28/3

(Happy) - September 21st 2010, 01:17 PMSoroban
Hello, statman101!

The cross-product approach is the best way.

Here's another (primitive and longer) approach.

(Use it when you're desperate.)

Quote:

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. - September 22nd 2010, 03:46 AMHallsofIvy