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To find the angle between the planes \( 2x - 6y - z = 1 \) and \( 3x - y + 4z = 6 \), we can use the normal vectors of each plane. The angle between the planes is the same as the angle between these normal vectors.

### Step 1: Identify the Normal Vectors
For the plane \( 2x - 6y - z = 1 \):
- The normal vector \( \vec{n}_1 = \langle 2, -6, -1 \rangle \).

For the plane \( 3x - y + 4z = 6 \):
- The normal vector \( \vec{n}_2 = \langle 3, -1, 4 \rangle \).

### Step 2: Use the Dot Product to Find the Cosine of the Angle
The formula for the angle \( \theta \) between two vectors \( \vec{a} \) and \( \vec{b} \) is:
\[
\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}
\]
where \( \vec{a} \cdot \vec{b} \) is the dot product of the vectors, and \( |\vec{a}| \) and \( |\vec{b}| \) are the magnitudes of the vectors.

#### Calculate the Dot Product \( \vec{n}_1 \cdot \vec{n}_2 \):
\[
\vec{n}_1 \cdot \vec{n}_2 = (2)(3) + (-6)(-1) + (-1)(4) = 6 + 6 - 4 = 8
\]

#### Calculate the Magnitudes \( |\vec{n}_1| \) and \( |\vec{n}_2| \):
\[
|\vec{n}_1| = \sqrt{2^2 + (-6)^2 + (-1)^2} = \sqrt{4 + 36 + 1} = \sqrt{41}
\]
\[
|\vec{n}_2| = \sqrt{3^2 + (-1)^2 + 4^2} = \sqrt{9 + 1 + 16} = \sqrt{26}
\]

### Step 3: Substitute into the Cosine Formula
\[
\cos \theta = \frac{8}{\sqrt{41} \cdot \sqrt{26}}
\]
Calculating this value:

1. First, find \( \sqrt{41} \approx 6.4 \) and \( \sqrt{26} \approx 5.1 \).
2. Multiply these values: \( \sqrt{41} \cdot \sqrt{26} \approx 6.4 \cdot 5.1 = 32.64 \).
3. Divide to find \( \cos \theta \): \( \cos \theta \approx \frac{8}{32.64} \approx 0.245 \).

### Step 4: Find \( \theta \) by Taking the Arccosine
\[
\theta = \cos^{-1}(0.245)
\]
Using a calculator, \( \theta \approx 75.8^\circ \).

### Final Answer
The angle between the planes is approximately \( 75.8^\circ \), rounded to the nearest tenth.

Would you like more details or have any questions?

### Related Questions
1. How do you determine the normal vector of a plane?
2. What is the geometric interpretation of the angle between two planes?
3. How can the dot product help in determining angles between vectors?
4. Why do we use the magnitudes of vectors in the cosine formula?
5. How would the process change if we were asked for the smallest angle between the planes?

#### Tip
In three-dimensional geometry, the angle between two planes is a measure of how “tilted” they are relative to each other. The closer this angle is to 90 degrees, the more perpendicular they are.

Find the angle between the planes 2x - 6y - z = 1 and 3x - y + 4z = 6. Round your answer to the nearest tenth of a degree.

Learn how to find the angle between two planes given by the equations 2x - 6y - z = 1 and 3x - y + 4z = 6. This problem demonstrates the use of normal vectors and the dot product to calculate the angle between planes, ideal for college-level students studying vectors and plane geometry.

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