Question

Find an equation of a plane containing the line r=⟨0,4,4⟩+t⟨−3,−2,1⟩ which is parallel to the plane 1x−1y+1z=−5 in which the coefficient of x is 1.

..?.. = 0.

Answer 1

The plane you want is parallel to another plane, x - y + z = -5, so they share a normal vector. In this case, it's ⟨1, -1, 1⟩.

The plane must also pass through the point (0, 4, 4) since it contains r(t). Then the equation of the plane is

⟨x, y - 4, z - 4⟩ • ⟨1, -1, 1⟩ = 0

x - (y - 4) + (z - 4) = 0

x - y + z = 0