Consider the following planes.
x + y + z = 6,     x + 8y + 8z = 6
(a)
Find parametric equations for the line of intersection of the planes. (Use the parameter t.)

x(t), y(t), z(t)

=

Correct: Your answer is correct.

(b)
Find the angle, in degrees rounded to one decimal place, between the planes.

### (a) Parametric Equations for the Line of Intersection:

To find the parametric equations for the line of intersection of the two planes, we need to solve the system of equations:
1. \( x + y + z = 6 \)
2. \( x + 8y + 8z = 6 \)

Step 1: Subtract the first equation from the second to eliminate \(x\):
\[
(x + 8y + 8z) - (x + y + z) = 6 - 6
\]
\[
7y + 7z = 0
\]
This simplifies to:
\[
y + z = 0 \quad \text{or} \quad z = -y
\]

Step 2: Substitute \(z = -y\) into the first equation:
\[
x + y + (-y) = 6
\]
\[
x = 6
\]

So, we now know \( x = 6 \) and \( z = -y \). We can parametrize \( y \) as \( y = t \) and substitute into the equations:
\[
x(t) = 6, \quad y(t) = t, \quad z(t) = -t
\]

Thus, the parametric equations for the line of intersection are:
\[
x(t) = 6, \quad y(t) = t, \quad z(t) = -t
\]

---

### (b) Angle Between the Planes:

The angle \( \theta \) between two planes can be found using the formula:
\[
\cos \theta = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{|\mathbf{n_1}| |\mathbf{n_2}|}
\]
where \( \mathbf{n_1} \) and \( \mathbf{n_2} \) are the normal vectors of the planes.

The normal vectors of the two planes are:
\[
\mathbf{n_1} = \langle 1, 1, 1 \rangle, \quad \mathbf{n_2} = \langle 1, 8, 8 \rangle
\]

Step 1: Compute the dot product \( \mathbf{n_1} \cdot \mathbf{n_2} \):
\[
\mathbf{n_1} \cdot \mathbf{n_2} = (1)(1) + (1)(8) + (1)(8) = 1 + 8 + 8 = 17
\]

Step 2: Find the magnitudes \( |\mathbf{n_1}| \) and \( |\mathbf{n_2}| \):
\[
|\mathbf{n_1}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}, \quad |\mathbf{n_2}| = \sqrt{1^2 + 8^2 + 8^2} = \sqrt{129}
\]

Step 3: Use the formula for \( \cos \theta \):
\[
\cos \theta = \frac{17}{\sqrt{3} \cdot \sqrt{129}} = \frac{17}{\sqrt{387}}
\]

Step 4: Find the angle \( \theta \) by taking the inverse cosine:
\[
\theta = \cos^{-1}\left( \frac{17}{\sqrt{387}} \right)
\]

Let me calculate this expression.

The angle between the planes is approximately \( 30.2^\circ \).

Would you like further details or clarifications?

Here are 5 related questions to explore this concept further:
1. How can we find the distance between two parallel planes?
2. What happens if two planes do not intersect? How can we check if they are parallel?
3. How do parametric equations represent a line in three-dimensional space?
4. How can we determine if the angle between two planes is \( 90^\circ \) (i.e., they are perpendicular)?
5. What is the significance of the dot product in finding angles between vectors?

**Tip:** The cross product of normal vectors can also be used to find the direction of the line of intersection of two planes.

Consider the following planes.
x + y + z = 6,     x + 8y + 8z = 6
(a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.)
(b) Find the angle, in degrees rounded to one decimal place, between the planes.

Learn how to find the parametric equations for the line of intersection of two planes and calculate the angle between them. This detailed solution covers linear algebra concepts such as parametric lines and the cosine formula for angles between vectors, suitable for Grades 11-12 and early college students.

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