Math Problem Statement

Find the acute angle between the planes 3x - y + 2z = 5 and x + 3y - 3z = 5. Enter your angle in radians to three significant figures.

Solution

To find the acute angle between the two planes 3xy+2z=53x - y + 2z = 5 and x+3y3z=5x + 3y - 3z = 5, we need to calculate the angle between their normal vectors.

Step 1: Identify the Normal Vectors

The general form of a plane equation is Ax+By+Cz=DAx + By + Cz = D, where n=A,B,C\vec{n} = \langle A, B, C \rangle is the normal vector.

For the planes given:

  • Plane 1: 3xy+2z=53x - y + 2z = 5 has a normal vector n1=3,1,2\vec{n_1} = \langle 3, -1, 2 \rangle.
  • Plane 2: x+3y3z=5x + 3y - 3z = 5 has a normal vector n2=1,3,3\vec{n_2} = \langle 1, 3, -3 \rangle.

Step 2: Use the Dot Product to Find the Angle

The angle θ\theta between two vectors n1\vec{n_1} and n2\vec{n_2} can be found using the dot product formula: cosθ=n1n2n1n2\cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|}

Calculate the Dot Product n1n2\vec{n_1} \cdot \vec{n_2}

n1n2=(3)(1)+(1)(3)+(2)(3)=336=6\vec{n_1} \cdot \vec{n_2} = (3)(1) + (-1)(3) + (2)(-3) = 3 - 3 - 6 = -6

Calculate the Magnitudes n1|\vec{n_1}| and n2|\vec{n_2}|

n1=32+(1)2+22=9+1+4=14|\vec{n_1}| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} n2=12+32+(3)2=1+9+9=19|\vec{n_2}| = \sqrt{1^2 + 3^2 + (-3)^2} = \sqrt{1 + 9 + 9} = \sqrt{19}

Substitute into the Cosine Formula

cosθ=61419\cos \theta = \frac{-6}{\sqrt{14} \cdot \sqrt{19}}

Calculate θ\theta

Using the inverse cosine function, we find: θ=cos1(61419)\theta = \cos^{-1} \left( \frac{-6}{\sqrt{14} \cdot \sqrt{19}} \right)

Using a calculator to approximate the value in radians (rounded to three significant figures): θ1.89 radians\theta \approx 1.89 \text{ radians}

So, the acute angle between the planes is approximately 1.89 radians.


Would you like further details on any part of this calculation?

Related Questions

  1. How do you find the angle between two lines instead of planes?
  2. What is the geometric interpretation of the dot product in finding angles?
  3. How would you calculate angles between planes in 4-dimensional space?
  4. Can the angle between planes be larger than 90 degrees?
  5. How would the solution change if the planes were parallel?

Tip

When working with planes, always verify the direction of normal vectors as they influence the orientation and the resulting angle between the planes.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Geometry
Plane Equations

Formulas

Dot Product: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \)
Magnitude of a Vector: \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \)

Theorems

Dot Product Theorem

Suitable Grade Level

Undergraduate