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To find the acute angle between two planes, we can use the following method:

1. The equation of a plane is given in the form:  
   \( ax + by + cz = d \),  
   where \( \mathbf{n} = \langle a, b, c \rangle \) is the normal vector to the plane.

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

### Step 1: Identify the normal vectors
For the planes:
1. \( 3x + 2y + 5z = 6 \), the normal vector is \( \mathbf{n_1} = \langle 3, 2, 5 \rangle \).
2. \( x - 3y - z = 5 \), the normal vector is \( \mathbf{n_2} = \langle 1, -3, -1 \rangle \).

### Step 2: Dot product of the normal vectors
The dot product \( \mathbf{n_1} \cdot \mathbf{n_2} \) is:
\[
\mathbf{n_1} \cdot \mathbf{n_2} = (3)(1) + (2)(-3) + (5)(-1) = 3 - 6 - 5 = -8
\]

### Step 3: Magnitudes of the normal vectors
The magnitudes of \( \mathbf{n_1} \) and \( \mathbf{n_2} \) are:
\[
|\mathbf{n_1}| = \sqrt{3^2 + 2^2 + 5^2} = \sqrt{9 + 4 + 25} = \sqrt{38}
\]
\[
|\mathbf{n_2}| = \sqrt{1^2 + (-3)^2 + (-1)^2} = \sqrt{1 + 9 + 1} = \sqrt{11}
\]

### Step 4: Calculate the cosine of the angle
Now, we can calculate \( \cos \theta \):
\[
\cos \theta = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{|\mathbf{n_1}| |\mathbf{n_2}|} = \frac{|-8|}{\sqrt{38} \cdot \sqrt{11}} = \frac{8}{\sqrt{418}}
\]
\[
\cos \theta = \frac{8}{20.46} \approx 0.391
\]

### Step 5: Find the angle \( \theta \)
To find \( \theta \), take the inverse cosine:
\[
\theta = \cos^{-1}(0.391) \approx 1.17 \text{ radians}
\]

So, the acute angle between the planes is approximately **1.17 radians**.

Let me know if you want more details or have any questions! Here are some related questions to expand on this topic:

1. How do we calculate the dot product of vectors?
2. What does the magnitude of a vector represent in geometry?
3. How do we convert radians to degrees?
4. Can two planes be parallel and what would their angle be?
5. What happens if the angle between two planes is obtuse?

**Tip**: Always check whether the angle is acute (between 0 and \( \pi/2 \)) when using dot products in geometry problems!

Use a calculator to find the acute angle between the planes: 3x + 2y + 5z = 6 and x − 3y − z = 5. (Round to two decimal places.)

Calculate the acute angle between the planes 3x + 2y + 5z = 6 and x − 3y − z = 5 using the dot product formula and vector magnitudes. This geometry problem demonstrates how to find angles between planes, suitable for high school students in Grades 10-12.

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