In analytical geometry, given the equations of two intersecting planes in 3-space, how can I find the equation of the line that their intersection forms?
The equation of a line in 3-space can be written as

L = P + tV,

where P is a point on the line, V is the direction vector of the line, and t is a real number.

To find the equation of the line that is formed by the intersection of two planes, we need to first find a point that lies on the line of intersection. This can be done by solving the two plane equations simultaneously. Once we have a point that lies on the line, we need to find the direction vector of the line. This can be done by finding the cross product of the two plane normal vectors. Finally, we can plug the point P and the direction vector V into the equation of the line above.

Let's say we have the following equations for our two planes:

P1: 2x + 3y - 5z = 10 
P2: 4x - y + 2z = 8

To find a point that lies on the line of intersection, we can solve the two plane equations simultaneously. We can do this by subtracting the two equations to get the following:
2x + 4y - 6z = 2

This equation can be rewritten in the form of x = ..., y = ..., and z = ..., where the right side of the equation is equal to 2. We can then solve for x, y, and z to get the point P = (1, -1, 1/3).

Now we need to find the direction vector of the line. We can do this by finding the cross product of the two plane normal vectors. The normal vectors of the two planes are:

P1: (2, 3, -5)
P2: (4, -1, 2)

The cross product of these two vectors is (14, -4, -18). This is the direction vector V of the line.

Finally, we can plug P and V into the equation of the line to get the equation of the line of intersection:

L = (1, -1, 1/3) + t(14, -4, -18)
Is there a version of this equation that doesn't require vectors?
Yes, there is a version of the equation of the line of intersection that doesn't require vectors. This version of the equation is called the parametric form of the line. In the parametric form, the equation of the line can be written as:

x = x1 + t(x2 - x1)
y = y1 + t(y2 - y1)
z = z1 + t(z2 - z1)

where (x1, y1, z1) and (x2, y2, z2) are two points on the line, and t is a real number.

To find two points on the line of intersection, we can use the two points that we found in the previous solution, one point from each plane. For example, we could use the point (1, -1, 1/3) and the point (3, 0, 1), which lies on both plane 1 and plane 2.

We can then plug these two points into the parametric form of the line to get the equation of the line of intersection:

x = 1 + t(3 - 1) = 1 + 2t
y = -1 + t(0 - (-1)) = -1 + t
z = 1/3 + t(1 - 1/3) = 1/3 + 2/3t

This is the equation of the line of intersection in parametric form, without the use of vectors.