Question

Consider the following planes. x + y + z = 4, x + 3y + 3z = 4?

(a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.)
(x(t), y(t), z(t)) =

(b) Find the angle between the planes. (Round your answer to one decimal place.)

Answer 1

(b)

Step 1:

The equations are

.

.

The normal vectors of two planes are

.

.

The angle between two planes is

The angle between two planes is .

Solution:

The angle between two planes is .

Comments

The correct angle is

7/sqrt(57) = cos(theta). Then take the inverse of cosine and then you'll have your angle.

22.0 degrees

Answer 2

(a)

Step 1:

The equations are

The cross product of two normal vectors is the direction vector for the line of intersection.

The normal vectors of two planes are

.

.

The cross product of two vectors is

The direction vector of line of intersection is .

Step 2:

Solve the eqn(1) and eqn(2).

Multiply eqn(1) on each side by 3.

.

Subtract eqn(2) from

Substitute in eqn(1)

.

There are no solutions for and .

So, take the point as .

The = is lie on the line of intersection.

The direction vector of line of intersection is .

The parametric plane equation is

The parametric equation is .

Solution:

The parametric equation is .