Find parametric equations for the line through (2, 4, 6) that is perpendicular to the plane x - y + 3z = 7.

Solution:

We know that

A vector which is perpendicular to the plane ax + by + cz + d = 0 is (a, b, c)

So the vector which is perpendicular to the plane x - y + 3z = 7 is (1, -1, 3)

A parametric equation of a line through (x₀, y₀, z₀) and parallel to vector (a, b, c) is

x = x₀ + t_a

y = y₀ + t_b

z = z₀ + t_c

The parametric equation of the line is

x = 2 + t

y = 4 - t

z = 6 + 3t

Therefore, the parametric equations of the line are x = 2 + t, y = 4 - t and z = 6 + 3t.

Find parametric equations for the line through (2, 4, 6) that is perpendicular to the plane x - y + 3z = 7.

Summary:

The parametric equations for the line through (2, 4, 6) that is perpendicular to the plane x - y + 3z = 7 are x = 2 + t, y = 4 - t and z = 6 + 3t.