Math Problem Statement

Find the acute angle (in degrees) between the planes -2y + z = -2 and -3x + 3y = -5.

Solution

To find the acute angle between two planes, we can use the formula for the angle between two planes in terms of the dot product of their normal vectors.

The angle θ\theta between two planes with normal vectors n1\mathbf{n}_1 and n2\mathbf{n}_2 is given by:

cosθ=n1n2n1n2\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{\|\mathbf{n}_1\| \|\mathbf{n}_2\|}

Step 1: Identify the normal vectors of the planes

The given equations of the planes are:

  1. 2y+z=2-2y + z = -2
  2. 3x+3y=5-3x + 3y = -5

For each plane, the coefficients of xx, yy, and zz give us the components of the normal vector:

  • For plane 1: n1=(0,2,1)\mathbf{n}_1 = (0, -2, 1)
  • For plane 2: n2=(3,3,0)\mathbf{n}_2 = (-3, 3, 0)

Step 2: Compute the dot product of the normal vectors

n1n2=(0)(3)+(2)(3)+(1)(0)=06+0=6\mathbf{n}_1 \cdot \mathbf{n}_2 = (0)(-3) + (-2)(3) + (1)(0) = 0 - 6 + 0 = -6

Step 3: Compute the magnitudes of the normal vectors

n1=02+(2)2+12=0+4+1=5\|\mathbf{n}_1\| = \sqrt{0^2 + (-2)^2 + 1^2} = \sqrt{0 + 4 + 1} = \sqrt{5}

n2=(3)2+32+02=9+9+0=18=32\|\mathbf{n}_2\| = \sqrt{(-3)^2 + 3^2 + 0^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}

Step 4: Compute the cosine of the angle

cosθ=n1n2n1n2=6532=6310=210\cos \theta = \frac{|\mathbf{n}_1 \cdot \mathbf{n}_2|}{\|\mathbf{n}_1\| \|\mathbf{n}_2\|} = \frac{| -6 |}{\sqrt{5} \cdot 3\sqrt{2}} = \frac{6}{3\sqrt{10}} = \frac{2}{\sqrt{10}}

Step 5: Find the acute angle in degrees

Now we find the angle θ\theta using the inverse cosine function:

θ=cos1(210)\theta = \cos^{-1}\left(\frac{2}{\sqrt{10}}\right)

Let's compute the angle in degrees.The acute angle between the planes 2y+z=2-2y + z = -2 and 3x+3y=5-3x + 3y = -5 is approximately 50.77 degrees.

Would you like further details on any of the steps, or do you have any additional questions?

Here are 5 related questions you might be interested in:

  1. How do you find the angle between two lines in 3D space?
  2. What is the geometric meaning of the dot product of two vectors?
  3. How do you find the angle between two planes if they are given in parametric form?
  4. How can you find the angle between a line and a plane?
  5. What is the relation between the cross product and the sine of the angle between two vectors?

Tip: When finding the angle between planes, ensure that the normal vectors are correctly identified from the coefficients of the plane equations.

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Math Problem Analysis

Mathematical Concepts

Vectors
Dot Product
Angles between Planes

Formulas

Angle between planes: cos(θ) = |n1 · n2| / (||n1|| ||n2||)
Dot Product: n1 · n2 = n1x * n2x + n1y * n2y + n1z * n2z
Magnitude of a vector: ||n|| = sqrt(x^2 + y^2 + z^2)

Theorems

Dot Product Theorem for Angles
Trigonometric Inverse Functions

Suitable Grade Level

Grades 11-12