find the angle between the two planes x + 6y − z = 6 and  6x − 6y + z = 1

To find the angle between two planes, we can use 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 two planes.
- \(\theta\) is the angle between the planes.

### Step 1: Identify the normal vectors of the two planes
The equation of a plane is generally written in the form:

\[
Ax + By + Cz = D
\]

Where \((A, B, C)\) is the normal vector of the plane.

For the two planes given:
1. \(x + 6y - z = 6\) has a normal vector \(\mathbf{n_1} = (1, 6, -1)\)
2. \(6x - 6y + z = 1\) has a normal vector \(\mathbf{n_2} = (6, -6, 1)\)

### Step 2: Calculate the dot product of \(\mathbf{n_1}\) and \(\mathbf{n_2}\)
The dot product of two vectors \((a_1, b_1, c_1)\) and \((a_2, b_2, c_2)\) is given by:

\[
\mathbf{n_1} \cdot \mathbf{n_2} = a_1a_2 + b_1b_2 + c_1c_2
\]

Substitute the values from the normal vectors:

\[
\mathbf{n_1} \cdot \mathbf{n_2} = (1)(6) + (6)(-6) + (-1)(1)
= 6 - 36 - 1 = -31
\]

### Step 3: Find the magnitudes of \(\mathbf{n_1}\) and \(\mathbf{n_2}\)
The magnitude of a vector \((a, b, c)\) is:

\[
|\mathbf{n}| = \sqrt{a^2 + b^2 + c^2}
\]

For \(\mathbf{n_1} = (1, 6, -1)\):

\[
|\mathbf{n_1}| = \sqrt{1^2 + 6^2 + (-1)^2} = \sqrt{1 + 36 + 1} = \sqrt{38}
\]

For \(\mathbf{n_2} = (6, -6, 1)\):

\[
|\mathbf{n_2}| = \sqrt{6^2 + (-6)^2 + 1^2} = \sqrt{36 + 36 + 1} = \sqrt{73}
\]

### Step 4: Calculate \(\cos \theta\)
Now we can use the formula:

\[
\cos \theta = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{|\mathbf{n_1}| |\mathbf{n_2}|}
= \frac{| -31 |}{\sqrt{38} \times \sqrt{73}} = \frac{31}{\sqrt{2774}}
\]

\[
\cos \theta = \frac{31}{52.66} \approx 0.5886
\]

### Step 5: Find the angle \(\theta\)
Now, take the inverse cosine to find the angle:

\[
\theta = \cos^{-1}(0.5886) \approx 54.00^\circ
\]

### Conclusion:
The angle between the two planes is approximately \(54.00^\circ\).

Would you like more details on any step, or do you have questions? Here are some related problems:

1. How do you find the distance between two parallel planes?
2. What is the geometric interpretation of the dot product?
3. How can you find the line of intersection between two planes?
4. How do you find the angle between two vectors in 3D?
5. How do you find the distance between a point and a plane?

**Tip**: The angle between two planes is the same as the angle between their normal vectors.

Find the angle between the two planes x + 6y − z = 6 and 6x − 6y + z = 1.

Learn how to calculate the angle between the two planes x + 6y − z = 6 and 6x − 6y + z = 1 using vector geometry. This solution covers the dot product formula, normal vectors, and the calculation of the angle between two planes in 3D space. Suitable for high school students in Grades 10-12.

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