Question

Find an equation of the plane.

The plane that contains the line
x = 3 + 2t,

y = t,

z = 6 − t
and is parallel to the plane
2x + 4y + 8z = 16

Answer 1

The normal to the given plane is (2, 4, 8), and the plane we want is parallel to this one so it has the same normal vector.

When t = 0, the given line, and thus the plane we want, passes through the point (3, 0, 6).

Then the equation of the plane is given by

(2, 4, 8) • (x - 3, y - 0, z - 6) = 0

2 (x - 3) + 4y + 8 (z - 6) = 0

2x + 4y + 8z = 54

or

x + 2y + 4z = 27

Answer 2

x +2y +4z = 27

Explanation:

The parallel plane will have the same coefficients of x, y, z as the given plane. We notice those have a common factor of 2, so the equation can be reduced to ...

x +2y +4z = constant

This equation is satisfied for every point on the line, so we have ...

(3 +2t) +2(t) +4(6 -t) = constant . . . . . substituting for x, y, z

3 +2t +2t +24 -4t = constant

27 = constant

The equation of the desired plane is ...

x +2y +4z = 27