Math Problem Statement

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.

Solution

To find the angle between the planes 2x6yz=12x - 6y - z = 1 and 3xy+4z=63x - 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 2x6yz=12x - 6y - z = 1:

  • The normal vector n1=2,6,1\vec{n}_1 = \langle 2, -6, -1 \rangle.

For the plane 3xy+4z=63x - y + 4z = 6:

  • The normal vector n2=3,1,4\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 θ\theta between two vectors a\vec{a} and b\vec{b} is: cosθ=abab\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} where ab\vec{a} \cdot \vec{b} is the dot product of the vectors, and a|\vec{a}| and b|\vec{b}| are the magnitudes of the vectors.

Calculate the Dot Product n1n2\vec{n}_1 \cdot \vec{n}_2:

n1n2=(2)(3)+(6)(1)+(1)(4)=6+64=8\vec{n}_1 \cdot \vec{n}_2 = (2)(3) + (-6)(-1) + (-1)(4) = 6 + 6 - 4 = 8

Calculate the Magnitudes n1|\vec{n}_1| and n2|\vec{n}_2|:

n1=22+(6)2+(1)2=4+36+1=41|\vec{n}_1| = \sqrt{2^2 + (-6)^2 + (-1)^2} = \sqrt{4 + 36 + 1} = \sqrt{41} n2=32+(1)2+42=9+1+16=26|\vec{n}_2| = \sqrt{3^2 + (-1)^2 + 4^2} = \sqrt{9 + 1 + 16} = \sqrt{26}

Step 3: Substitute into the Cosine Formula

cosθ=84126\cos \theta = \frac{8}{\sqrt{41} \cdot \sqrt{26}} Calculating this value:

  1. First, find 416.4\sqrt{41} \approx 6.4 and 265.1\sqrt{26} \approx 5.1.
  2. Multiply these values: 41266.45.1=32.64\sqrt{41} \cdot \sqrt{26} \approx 6.4 \cdot 5.1 = 32.64.
  3. Divide to find cosθ\cos \theta: cosθ832.640.245\cos \theta \approx \frac{8}{32.64} \approx 0.245.

Step 4: Find θ\theta by Taking the Arccosine

θ=cos1(0.245)\theta = \cos^{-1}(0.245) Using a calculator, θ75.8\theta \approx 75.8^\circ.

Final Answer

The angle between the planes is approximately 75.875.8^\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.

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Math Problem Analysis

Mathematical Concepts

Vectors
Angles between planes
Dot product
Normal vectors
Plane equations

Formulas

Cosine of angle formula: cos(θ) = (a · b) / (|a| |b|)
Magnitude of a vector: |a| = √(a₁² + a₂² + a₃²)
Dot product of vectors: a · b = a₁b₁ + a₂b₂ + a₃b₃

Theorems

Cosine angle formula between two vectors

Suitable Grade Level

College level (Calculus or Linear Algebra)