To find the parametric equations for the line that passes through the point (2, 4, 6) and is perpendicular to the plane defined by the equation x + y + 3z = 7, we need to determine the direction of the line and then use the point to create the equations.
The normal vector of the plane can be extracted from the coefficients of x, y, and z in the plane equation. For the given plane equation, the normal vector is N = (1, 1, 3).
This normal vector indicates the direction in which the line will travel since the line is perpendicular to the plane. We can then write the parametric equations for the line using the point and the direction vector.
The general form for the parametric equations of a line through a point (x_0, y_0, z_0) with a direction vector (a, b, c) is given by:
- x = x0 + at
- y = y0 + bt
- z = z0 + ct
Substituting the point (2, 4, 6) for (x_0, y_0, z_0) and the normal vector (1, 1, 3) for (a, b, c), we get:
- x = 2 + 1t
- y = 4 + 1t
- z = 6 + 3t
We can express this more neatly as:
- x = 2 + t
- y = 4 + t
- z = 6 + 3t
Thus, the parametric equations of the line through (2, 4, 6) that is perpendicular to the plane x + y + 3z = 7 are:
- x = 2 + t
- y = 4 + t
- z = 6 + 3t