Find a direction vector for the line of intersection of x+z=5 and x-y=3.?

Question

No need to show work but would appreciate it!

Answer 1

There are an infinite number of solutions since if you have 3 unknowns, you must have 3 equations.

Let x = t then z = 5 - t and y = t - 3

These are the only requirements.

Answer 2

Find a direction vector for the line of intersection of the planes x + z = 5 and x - y = 3.

The vector v, that is normal to the normal vectors of the planes will lie in both planes. It will be the directional vector of the line of intersection.

The plane x + z = 5 has normal vector n1 = <1, 0, 1>.
The plane x - y = 3 has normal vector n2 = <1, -1, 0>.

Take the cross product to find a vector normal to both. This will be the directional vector of the line of intersection.

v = n1 X n1 = <1, 0, 1> X <1, -1, 0> = <1, 1, -1>

Any non-zero multiple of v is also a directional vector of the line of intersection.