Problem 51 - Thomas Calculus

Question

Use a calculator to find the acute angles between the planes in Exercises 49–52 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 angle between the planes is approximately 0.93 radians.

Step 1: Identify Normal Vectors of the Planes

To find the angle between two planes, first identify their normal vectors. The normal vector for a plane with equation \(Ax + By + Cz = D\) is \( \langle A, B, C \rangle \). 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 Normal Vectors

The dot product of two vectors \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\) is calculated as \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\). For \(\mathbf{n_1} = \langle 2, 2, -1 \rangle\) and \(\mathbf{n_2} = \langle 1, 2, 1 \rangle\), the dot product is \(2 \times 1 + 2 \times 2 + (-1) \times 1 = 2 + 4 - 1 = 5\).

Step 3: Calculate the Magnitudes of Normal Vectors

The magnitude of a vector \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) is \(\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}\). For \(\mathbf{n_1} = \langle 2, 2, -1 \rangle\), the magnitude is \(\sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\). For \(\mathbf{n_2} = \langle 1, 2, 1 \rangle\), the magnitude is \(\sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\).

Step 4: Calculate the Cosine of the Angle

The cosine of the angle \(\theta\) between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by \(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\). Using the values calculated earlier, \(\cos \theta = \frac{5}{3 \cdot \sqrt{6}}\).

Step 5: Calculate the Angle Using Inverse Cosine

To find the angle \(\theta\), take the inverse cosine of the value found in the previous step. \(\theta = \cos^{-1}\left(\frac{5}{3\sqrt{6}}\right)\). Use a calculator to find \(\theta \approx 0.93\) radians, rounded to the nearest hundredth.

Key Concepts

Normal Vector

Every plane in 3D space is associated with a unique normal vector. This vector is perpendicular to the plane's surface and is a key ingredient in finding angles between planes. Normal vectors are derived from the plane's equation, typically written in the form \( Ax + By + Cz = D \).

• The coefficients \( A, B, \) and \( C \) form the components of the normal vector \( \langle A, B, C \rangle \).

For example, the plane \(2x + 2y - z = 3\) has a normal vector \( \mathbf{n_1} = \langle 2, 2, -1 \rangle \), and the plane \(x + 2y + z = 2\) has a normal vector \( \mathbf{n_2} = \langle 1, 2, 1 \rangle \). Understanding normal vectors helps visualize situations and solve problems involving plane-to-plane angles.

Dot Product

Once you have the normal vectors from the plane equations, the next step is calculating their dot product. The dot product is a numerical value that reveals the relationship between two vectors. To calculate it, multiply the corresponding components of the vectors and sum the results:

• For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

Using this, find the dot product of the normal vectors of the planes: \(2 \times 1 + 2 \times 2 + (-1) \times 1 = 5\). This value is crucial for determining the angle between the planes.

Magnitude of a Vector

The magnitude of a vector is akin to its length and is needed to find angles between vectors. It is a necessary component when computing the angle between vectors through their dot product. For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), its magnitude is calculated as:

• \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2} \)

For the normal vector \( \mathbf{n_1} = \langle 2, 2, -1 \rangle \), the magnitude is \( \sqrt{2^2 + 2^2 + (-1)^2} = 3 \). Similarly, the magnitude of \( \mathbf{n_2} = \langle 1, 2, 1 \rangle \) is \( \sqrt{6} \). These calculations provide the means to eventually determine the angle between the planes.

Inverse Cosine

The inverse cosine function, often written as \( \cos^{-1} \), allows us to calculate an angle from its cosine value. In the context of finding the angle between two planes, this is the final step. Once you have the dot product and vectors' magnitudes, compute the cosine of the angle \( \theta \) with:

• \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \)

For the normal vectors derived earlier, \( \cos \theta = \frac{5}{3 \cdot \sqrt{6}} \). Use inverse cosine to find \( \theta = \cos^{-1}\left(\frac{5}{3\sqrt{6}}\right) \), which evaluates to approximately 0.93 radians. This measurement ends the journey of determining the angle between the two planes.