Find the angle between the two planes. (Round your answer to two decimal places.) & Find a set of parametric equations for the line of intersection of the planes.

Question

Consider the following planes.

3x + 6y − z = 42

x − 12y + 2z = 0

(a) Find the angle between the two planes. (Round your answer to two decimal places.)

____ °

(b) Find a set of parametric equations for the line of intersection of the planes. (Use t for the parameter. Enter your answers as a comma-separated list of equations.)

Answer 1

Alternate Answer using Cross Product for part (b):

(a) The angle between the two planes is the angle between the normal vectors which define those planes (maybe take a moment to convince yourself of this).

The normal vectors for the planes are 〈3, 6, -1〉 and 〈1, -12, 2〉 from the equations above. Whenever we have vectors and are asked about the angle between them, we should try to utilize either the dot product or cross product. In this case, we'll use the dot product since it's easier to compute.

Recall if u =〈u₁, u₂, u₃〉and v = 〈v₁, v₂, v₃〉:

u • v = |u| |v| cos(θ) and u • v = u₁v₁ + u₂v₂ + u₃v₃

With our vectors:

3(1) + 6(-12) + (-1)(2) = sqrt(46) · sqrt(149) cos(θ).

cos(θ) = -71 / sqrt(6854)

θ ≈ 149º. Another acceptable answer would be the supplementary angle to this: θ ≈ 31º.

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(b) To write the parametric equations for the line of intersection between the planes, we need a point on the line and a vector parallel to the line. Once we find these, we can write the vector equation as

r(t) = r₀ + tv ; where r₀ is the vector from the origin to the point and v is the vector parallel to the line.

To find a point: The equations of the planes give us a system of equations with a free variable (we can choose whatever value we like for any one of the variables before the other two are locked into place) since we have 3 variables and only 2 equations. Let's choose, arbitrarily, z = 0. Then x = 12 and y = 1 and thus we have found a point on the line of intersection (12, 1, 0).

To find a vector parallel to the line: Notice that the line of intersection lies in both planes and thus is perpendicular to the normal vector of each plane. Restated: we're looking for a vector that is perpendicular to two other vectors. Recall that the resultant vector of the cross product of 2 vectors u and v is perpendicular to both u and v. So if we take the cross product of the normal vectors of the planes, we'll get a vector that is perpendicular to both and thus parallel to the line of intersection. Doing so:

〈3, 6, -1〉×〈1, -12, 2〉= 〈0, -7, -42〉 this is the vector parallel to the intersection line.

The vector equation of the line is therefore:

r(t) = 〈12, 1, 0〉+〈0, -7, -42〉t

and the parametric equations follow from the components of our vector equation:

x = 12

y = 1 - 7t

z = -42t

where t ∈ ℝ.