Determine an equation for the line parallel to 2x + 6y + 4z = 1 and contains the point P(3, 2, 1).
To find an equation for the line parallel to a given plane and passing through a given point, we need to determine the direction of the line.
First, let's find the normal vector of the given plane. The equation of the plane is 2x + 6y + 4z = 1. The coefficients of x, y, and z in this equation represent the components of the normal vector. In this case, the normal vector is (2, 6, 4).
Since the line we're looking for is parallel to the plane, it will have the same direction as the normal vector.
Now we have the direction vector, but we need a specific point on the line. We're given the point P (3, 2, 1), which lies on the line.
Using these pieces of information, we can write the equation for the line. The general equation for a line in vector form is:
r = a + t * b
where r is a position vector, a is a known position vector on the line, t is a scalar parameter, and b is the direction vector of the line.
In our case, the equation becomes:
r = (3, 2, 1) + t * (2, 6, 4)
This equation represents the line parallel to the plane 2x + 6y + 4z = 1 that passes through the point P(3, 2, 1).