Problem 55 - University Calculus: Early Transcendentals

Question

Use a calculator to find the acute angles between the planes to the nearest hundredth of a radian. $$2 x+2 y-z=3, \quad x+2 y+z=2$$

Step-by-Step Solution

Answer

The acute angle between the planes is approximately 0.39 radians.

Step 1: Identify the Normal Vectors

The normal vector of the plane is derived from the coefficients of the variables in the plane equation. For the plane \(2x + 2y - z = 3\), the normal vector is \(\mathbf{n_1} = \langle 2, 2, -1 \rangle\). For the plane \(x + 2y + z = 2\), the normal vector is \(\mathbf{n_2} = \langle 1, 2, 1 \rangle\).

Step 2: Calculate the Dot Product of the Normals

The dot product \(\mathbf{n_1} \cdot \mathbf{n_2}\) is calculated using the formula: \(2 \cdot 1 + 2 \cdot 2 + (-1) \cdot 1 = 2 + 4 - 1 = 5\).

Step 3: Calculate the Magnitudes of Each Normal Vector

Compute the magnitude of \(\mathbf{n_1}\) as \(\|\mathbf{n_1}\| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{9} = 3\). For \(\mathbf{n_2}\), the magnitude is \(\|\mathbf{n_2}\| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}\).

Step 4: Use Dot Product to Find Cosine of the Angle

The cosine of the angle \(\theta\) between the planes is \(\cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{\|\mathbf{n_1}\| \|\mathbf{n_2}\|}\). Substitute the values: \(\cos \theta = \frac{5}{3\sqrt{6}}\).

Step 5: Calculate the Angle and Convert to Radians

Find the angle \(\theta\) by taking the inverse cosine of \(\frac{5}{3\sqrt{6}}\). Use the calculator to compute \(\theta = \cos^{-1}\left(\frac{5}{3\sqrt{6}}\right) \approx 0.39\) radians to the nearest hundredth.

Key Concepts

Normal Vector

A normal vector is a key concept in plane geometry, representing a vector that is perpendicular to a plane or surface. Understanding how to find and utilize normal vectors is important for solving plane-related problems.
For a given plane with the equation in the form of \(ax + by + cz = d\), the normal vector is derived from the coefficients \(a\), \(b\), and \(c\).
In simpler terms:

• The **normal vector** \(\mathbf{n}\) = \(\langle a, b, c \rangle\) signifies a direction perpendicular to the plane.
• It is critical for determining angles between planes and projecting components.

Let's consider the exercise example. For the plane equation \(2x + 2y - z = 3\), the corresponding normal vector \(\mathbf{n_1}\) is obtained by looking at the coefficients resulting in \(\langle 2, 2, -1 \rangle\).
Similarly, for the plane \(x + 2y + z = 2\), the normal vector \(\mathbf{n_2}\) would be \(\langle 1, 2, 1 \rangle\). Understanding these vectors helps us move forward with calculating other properties between planes.

Dot Product

The dot product is a basic operation in vector algebra that combines two vectors, resulting in a scalar value. It is a crucial tool for plane geometry as it helps determine the angle between vectors, among other applications.
The formula for the dot product of vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\) is:

• \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\)

Essentially, it multiplies corresponding components of the vectors and adds the results.

In our example, the dot product of the normal vectors \(\mathbf{n_1} = \langle 2, 2, -1 \rangle\) and \(\mathbf{n_2} = \langle 1, 2, 1 \rangle\) is calculated as follows:

• \(\mathbf{n_1} \cdot \mathbf{n_2} = 2 \cdot 1 + 2 \cdot 2 + (-1) \cdot 1 = 5\)

This operation is a powerful way to uncover deeper geometric relationships between planes and vectors.

Angle Between Planes

Understanding the angle between planes involves determining how two planes are oriented relative to each other. This is achieved by looking at their normal vectors and using the dot product to find the cosine of the angle.
Once we have the dot product from the normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\), we utilize the formula:

• \(\cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{\|\mathbf{n_1}\| \|\mathbf{n_2}\|}\)

This formula relates the dot product to the magnitudes of the normal vectors to find the cosine of the angle \(\theta\) between them.

In the step-by-step solution, substituting the known values gives us:

• The dot product is \(5\), and the magnitudes are \(3\) and \(\sqrt{6}\), leading to \(\cos \theta = \frac{5}{3\sqrt{6}}\).

Solving for \(\theta\) involves calculating the inverse cosine \(\cos^{-1}\) of this ratio, resulting in approximately \(0.39\) radians.
This process helps us precisely define the spatial relationship between the two geometrical planes.

Magnitude of a Vector

The magnitude of a vector is its length and denotes how "strong" or "intense" the vector is. For a vector \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\), its magnitude \(\|\mathbf{v}\|\) is found using this formula:

• \(\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}\)

This essentially applies the Pythagorean Theorem in three dimensions.

In our exercise, the magnitude of normal vector \(\mathbf{n_1} = \langle 2, 2, -1 \rangle\) is:

• \(\|\mathbf{n_1}\| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{9} = 3\)

Similarly, the magnitude for \(\mathbf{n_2} = \langle 1, 2, 1 \rangle\) is:

• \(\|\mathbf{n_2}\| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}\)

Calculating magnitudes provides essential information for understanding and solving geometric problems, particularly in assessing angles and vectors' relative sizes.