Problem 49 - 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. $$ 2 x+2 y+2 z=3, \quad 2 x-2 y-z=5 $$

Step-by-Step Solution

Answer

The acute angle between the planes is approximately 1.23 radians.

Step 1: Identify the Normal Vectors

Each plane can be represented by a linear equation of the form \(Ax + By + Cz = D\). The normal vector is \([A, B, C]\). For the first plane, \(2x + 2y + 2z = 3\), the normal vector is \(\mathbf{n_1} = [2, 2, 2]\). For the second plane, \(2x - 2y - z = 5\), the normal vector is \(\mathbf{n_2} = [2, -2, -1]\).

Step 2: Formula for Angle Between Two Vectors

The formula for finding the angle \(\theta\) between two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by the dot product: \[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}.\]

Step 3: Compute the Dot Product

Calculate the dot product of the normal vectors: \(\mathbf{n_1} \cdot \mathbf{n_2} = 2 \times 2 + 2 \times (-2) + 2 \times (-1) = 4 - 4 - 2 = -2\).

Step 4: Compute Magnitudes of Normal Vectors

Calculate the magnitudes of the vectors: \(\|\mathbf{n_1}\| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}\), and \(\|\mathbf{n_2}\| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3\).

Step 5: Substitute into the Cosine Formula

Substitute the dot product and magnitudes into the cosine formula: \[\cos \theta = \frac{-2}{2\sqrt{3} \times 3} = \frac{-2}{6\sqrt{3}} = \frac{-1}{3\sqrt{3}}.\]

Step 6: Compute the Angle

Use a calculator to find \(\theta = \cos^{-1}\left(\frac{-1}{3\sqrt{3}}\right)\). Compute the angle in radians to the nearest hundredth: \(\theta \approx 1.23\) radians.

Key Concepts

Normal Vectors

When dealing with planes in three-dimensional space, normal vectors are crucial. A normal vector is a vector that is perpendicular to the plane. You can find it easily from any plane's equation, like the standard form equation, \(Ax + By + Cz = D\). Here, the normal vector is simply \([A, B, C]\).
For the exercise at hand:

• The first plane given is \(2x + 2y + 2z = 3\). Therefore, its normal vector is \(\mathbf{n_1} = [2, 2, 2]\).
• The second plane is \(2x - 2y - z = 5\). Its normal vector is \(\mathbf{n_2} = [2, -2, -1]\).

Understanding these normal vectors allows you to project the planes' characteristics into the realm of vector operations. This aids in further calculations, like finding angles between planes by using their normal vectors.

Dot Product

The dot product is a fundamental operation in vector algebra. It gives us a way to multiply two vectors and obtain a scalar (a single number). For two vectors, \(\mathbf{a} = [a_1, a_2, a_3]\) and \(\mathbf{b} = [b_1, b_2, b_3]\), the dot product is computed as:\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3\]In our particular exercise, the dot product of the normal vectors, \(\mathbf{n_1}\) and \(\mathbf{n_2}\), is:\[\mathbf{n_1} \cdot \mathbf{n_2} = 2 \times 2 + 2 \times (-2) + 2 \times (-1) = 4 - 4 - 2 = -2\]This result, \(-2\), will play a vital role in calculating the angle between the planes. The dot product helps determine the cosine of the angle.

Magnitude of Vectors

To find the angle between planes, we also need to know the magnitude of the normal vectors involved. The magnitude of a vector is its length and is calculated using the square root of the sum of its components squared. For a vector \(\mathbf{a} = [a_1, a_2, a_3]\), the magnitude \(\|\mathbf{a}\|\) is given by:\[\|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2 + a_3^2}\]Applying this to our vectors:

• The magnitude of \(\mathbf{n_1} = [2, 2, 2]\) is \(\|\mathbf{n_1}\| = \sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}\).
• The magnitude of \(\mathbf{n_2} = [2, -2, -1]\) is \(\|\mathbf{n_2}\| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3\).

With these magnitudes, we can now use them alongside the dot product to determine the angle.

Cosine Formula for Angles

Finding the angle between two vectors, like the normal vectors of two planes, involves the cosine formula. This formula relates the dot product of the vectors with their magnitudes. The formula is:\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\]For the normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\), we calculated the dot product as \(-2\), and their magnitudes as \(2\sqrt{3}\) and \(3\) respectively. Substituting these values into the cosine formula, we get:\[\cos \theta = \frac{-2}{2\sqrt{3} \times 3} = \frac{-2}{6\sqrt{3}} = \frac{-1}{3\sqrt{3}}\]This expression gives the cosine of the angle, \(\theta\), between the planes. To find \(\theta\) itself, use a calculator to compute the inverse cosine (\(\cos^{-1}\)), rounding the result to the nearest hundredth of a radian, which results in approximately \(1.23\) radians.