Math Problem Statement

Consider the following planes. x + y + z = 6, x + 8y + 8z = 6 (a) Find parametric equations for the line of intersection of the planes. (Use the parameter t.)

x(t), y(t), z(t)

=

Correct: Your answer is correct.

(b) Find the angle, in degrees rounded to one decimal place, between the planes.

Solution

(a) Parametric Equations for the Line of Intersection:

To find the parametric equations for the line of intersection of the two planes, we need to solve the system of equations:

  1. x+y+z=6x + y + z = 6
  2. x+8y+8z=6x + 8y + 8z = 6

Step 1: Subtract the first equation from the second to eliminate xx: (x+8y+8z)(x+y+z)=66(x + 8y + 8z) - (x + y + z) = 6 - 6 7y+7z=07y + 7z = 0 This simplifies to: y+z=0orz=yy + z = 0 \quad \text{or} \quad z = -y

Step 2: Substitute z=yz = -y into the first equation: x+y+(y)=6x + y + (-y) = 6 x=6x = 6

So, we now know x=6x = 6 and z=yz = -y. We can parametrize yy as y=ty = t and substitute into the equations: x(t)=6,y(t)=t,z(t)=tx(t) = 6, \quad y(t) = t, \quad z(t) = -t

Thus, the parametric equations for the line of intersection are: x(t)=6,y(t)=t,z(t)=tx(t) = 6, \quad y(t) = t, \quad z(t) = -t


(b) Angle Between the Planes:

The angle θ\theta between two planes can be found using 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 planes.

The normal vectors of the two planes are: n1=1,1,1,n2=1,8,8\mathbf{n_1} = \langle 1, 1, 1 \rangle, \quad \mathbf{n_2} = \langle 1, 8, 8 \rangle

Step 1: Compute the dot product n1n2\mathbf{n_1} \cdot \mathbf{n_2}: n1n2=(1)(1)+(1)(8)+(1)(8)=1+8+8=17\mathbf{n_1} \cdot \mathbf{n_2} = (1)(1) + (1)(8) + (1)(8) = 1 + 8 + 8 = 17

Step 2: Find the magnitudes n1|\mathbf{n_1}| and n2|\mathbf{n_2}|: n1=12+12+12=3,n2=12+82+82=129|\mathbf{n_1}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}, \quad |\mathbf{n_2}| = \sqrt{1^2 + 8^2 + 8^2} = \sqrt{129}

Step 3: Use the formula for cosθ\cos \theta: cosθ=173129=17387\cos \theta = \frac{17}{\sqrt{3} \cdot \sqrt{129}} = \frac{17}{\sqrt{387}}

Step 4: Find the angle θ\theta by taking the inverse cosine: θ=cos1(17387)\theta = \cos^{-1}\left( \frac{17}{\sqrt{387}} \right)

Let me calculate this expression.

The angle between the planes is approximately 30.230.2^\circ.

Would you like further details or clarifications?

Here are 5 related questions to explore this concept further:

  1. How can we find the distance between two parallel planes?
  2. What happens if two planes do not intersect? How can we check if they are parallel?
  3. How do parametric equations represent a line in three-dimensional space?
  4. How can we determine if the angle between two planes is 9090^\circ (i.e., they are perpendicular)?
  5. What is the significance of the dot product in finding angles between vectors?

Tip: The cross product of normal vectors can also be used to find the direction of the line of intersection of two planes.

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Math Problem Analysis

Mathematical Concepts

Linear Algebra
Geometry
Planes in 3D

Formulas

Parametric equation of a line: x(t), y(t), z(t)
Cosine of the angle between planes: cos(θ) = |n1 · n2| / (|n1| |n2|)

Theorems

Dot Product
Angle between vectors

Suitable Grade Level

Grades 11-12, Early College