Question

(a) Find a vector parallel to the line of intersection of the planes −4x+2y−z=1 and 3x−2y+2z=1.

Answer 1

Find the intersection of the two planes. Do this by solving for z in terms of x and y ; then solve for y in terms of x ; then again for z but only in terms of x.

-4x + 2y - z = 1   ==>   z = -4x + 2y - 1

3x - 2y + 2z = 1   ==>   z = (1 - 3x + 2y)/2

==>   -4x + 2y - 1 = (1 - 3x + 2y)/2

==>   -8x + 4y - 2 = 1 - 3x + 2y

==>   -5x + 2y = 3

==>   y = (3 + 5x)/2

==>   z = -4x + 2 (3 + 5x)/2 - 1 = x + 2

So if we take x = t, the line of intersection is parameterized by

r(t) = ⟨t, (3 + 5t )/2, 2 + t⟩

Just to not have to work with fractions, scale this by a factor of 2, so that

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩

(a) The tangent vector to r(t) is parallel to this line, so you can use

v = dr/dt = d/dt ⟨2t, 3 + 5t, 4 + 2t⟩ = ⟨2, 5, 2⟩

or any scalar multiple of this.

(b) (-1, -1, 1) indeed lies in both planes. Plug in x = -1, y = 1, and z = 1 to both plane equations to see this for yourself. We already found the parameterization for the intersection,

r(t) = ⟨2t, 3 + 5t, 4 + 2t⟩