Math Problem Statement

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

Solution

To find the angle between two planes, we can use the formula:

cosθ=n1n2n1n2\cos \theta = \frac{|\mathbf{n_1} \cdot \mathbf{n_2}|}{|\mathbf{n_1}| |\mathbf{n_2}|}

Where:

  • n1\mathbf{n_1} and n2\mathbf{n_2} are the normal vectors of the two planes.
  • θ\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=DAx + By + Cz = D

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

For the two planes given:

  1. x+6yz=6x + 6y - z = 6 has a normal vector n1=(1,6,1)\mathbf{n_1} = (1, 6, -1)
  2. 6x6y+z=16x - 6y + z = 1 has a normal vector n2=(6,6,1)\mathbf{n_2} = (6, -6, 1)

Step 2: Calculate the dot product of n1\mathbf{n_1} and n2\mathbf{n_2}

The dot product of two vectors (a1,b1,c1)(a_1, b_1, c_1) and (a2,b2,c2)(a_2, b_2, c_2) is given by:

n1n2=a1a2+b1b2+c1c2\mathbf{n_1} \cdot \mathbf{n_2} = a_1a_2 + b_1b_2 + c_1c_2

Substitute the values from the normal vectors:

= 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.

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Math Problem Analysis

Mathematical Concepts

Vector Geometry
Dot Product
Angle Between Planes

Formulas

cos(θ) = |n1 • n2| / (|n1| |n2|)
Dot product: n1 • n2 = a1a2 + b1b2 + c1c2
Magnitude of a vector: |n| = √(a² + b² + c²)

Theorems

Dot Product Theorem
Plane Equation in 3D

Suitable Grade Level

Grades 10-12