Final answer:
The symmetric equations for the line of intersection of the given planes z=3x-y-10 and z=5x+2y-13 are derived by setting the z components equal, solving for y in terms of x, and then finding z in terms of the parameter t. A point on the line is then used to express the line symmetrically.
Step-by-step explanation:
When finding symmetric equations for the line of intersection between two planes, we must first determine a direction vector for the line and a point through which the line passes. These planes are given as z=3x−y−10 and z=5x+2y−13. By setting these equations equal to each other, since any point on the line of intersection must satisfy both plane equations, we can find a relationship between x and y.
Substitute z from one equation into the other, yielding 3x - y - 10 = 5x + 2y - 13. Simplifying, we get -2x + 3y = 3. This is a linear equation in x and y. We can express x or y in terms of the other variable to get a directional vector. Let's set x = t (a parameter), and then from -2t + 3y = 3, we get y = (2t + 3)/3.
To find z in terms of t, we can substitute x = t into either of the original plane equations. Using z = 3x - y - 10, we get z = 3t - ((2t + 3)/3) - 10 which simplifies to z = (7t - 33)/3.
We need a point on the line of intersection for our symmetric equations. We can pick t = 0 for simplicity, which gives us x = 0, y = 1, and z = -11. So one point on the line is (0, 1, -11).
The symmetric equations of the line can now be given by the point and the directional vector components we derived: x = 0 + t, y = 1 + (2t + 3)/3, and z = -11 + (7t - 33)/3. To make these more symmetric, solve for the parameter t in each equation: t = x, t = (3y - 3)/2, and t = (3z + 33)/7.
Hence, the symmetric equations for the line of intersection are x = t, y = (2t + 3)/3, and z = (7t - 33)/3.