Problem 47 - University Calculus: Early Transcendentals

Question

Find the angles between the planes. $$x+y=1, \quad 2 x+y-2 z=2$$

Step-by-Step Solution

Answer

The angle between the planes is \(45^\circ\) or \(\frac{\pi}{4}\) radians.

Step 1: Understand the Problem

We need to find the angle between two planes represented by their equations: \(x + y = 1\) and \(2x + y - 2z = 2\). The angle between two planes is the angle between their normal vectors.

Step 2: Identify the Normal Vectors

The normal vector to the plane \(x+y=1\) can be extracted from its coefficients as \(\mathbf{n_1} = \langle 1, 1, 0 \rangle\). Similarly, the normal vector to the plane \(2x+y-2z=2\) is \(\mathbf{n_2} = \langle 2, 1, -2 \rangle\).

Step 3: Use the Dot Product Formula

The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \), where \(\theta\) is the angle between them. The dot product of \(\mathbf{n_1}\) and \(\mathbf{n_2}\) is found using \(1\times2 + 1\times1 + 0\times(-2) = 2 + 1 + 0 = 3\).

Step 4: Compute Magnitudes of the Normal Vectors

Calculate the magnitudes of \(\mathbf{n_1}\) and \(\mathbf{n_2}\). For \(\mathbf{n_1} = \langle 1, 1, 0 \rangle\), the magnitude is \(\sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}\). For \(\mathbf{n_2} = \langle 2, 1, -2 \rangle\), the magnitude is \(\sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3\).

Step 5: Substitute into the Dot Product Formula

Using the formula \(\cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{|\mathbf{n_1}| |\mathbf{n_2}|}\), substitute the values: \(\cos \theta = \frac{3}{\sqrt{2} \times 3}\). Simplifying, \(\cos \theta = \frac{1}{\sqrt{2}}\).

Step 6: Solve for \(\theta\)

To find \(\theta\), take the inverse cosine: \(\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\). This simplifies to \(\theta = 45^\circ\) or \(\frac{\pi}{4}\) radians, which is the angle between the planes.

Key Concepts

Dot Product

The dot product is a fundamental operation in vector mathematics, crucial for finding angles between vectors. It involves multiplying corresponding components of two vectors and summing these products. For example, if we have two 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 given by:

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

The result of the dot product is a scalar quantity, which is used to determine the cosine of the angle between the two vectors. The formula \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \) relates the dot product to the magnitudes of the vectors and the angle \( \theta \) between them. Understanding how to compute and use the dot product is essential when working with planes and their normal vectors.

Normal Vectors

Normal vectors play a significant role in geometry, especially when dealing with planes. A normal vector is a vector that is perpendicular to a surface. For a plane defined by an equation, its normal vector can be obtained from the coefficients of the variables in the equation.
For instance, in the equation \( x + y = 1 \), the normal vector \( \mathbf{n_1} = \langle 1, 1, 0 \rangle \) corresponds to the coefficients of \( x \) and \( y \). Similarly, for \( 2x + y - 2z = 2 \), the normal vector \( \mathbf{n_2} = \langle 2, 1, -2 \rangle \) comes from the coefficients of \( x, y, \) and \( z \).

• Normal vectors are crucial when determining the orientation of a plane in three-dimensional space.
• They allow us to calculate angles between planes by comparing their dot product and magnitudes.

Grasping the concept of normal vectors helps in solving a variety of geometric problems, including those involving intersections and angles.

Magnitude of Vectors

The magnitude, or length, of a vector is a measure of how long it is in space. It's calculated using the components of the vector. For a vector \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \), the magnitude is determined by:

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

In our problem, we calculated the magnitudes of the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \). For \( \mathbf{n_1} = \langle 1, 1, 0 \rangle \), the magnitude is \( \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2} \). For \( \mathbf{n_2} = \langle 2, 1, -2 \rangle \), the magnitude is \( \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3 \).
Understanding vector magnitude is fundamental in vector math, as it provides needed information to scale vectors, compute dot products, and find angles.

Inverse Cosine

The inverse cosine function, denoted as \( \cos^{-1} \), is used to find the angle when the cosine of that angle is known. It is a key step in calculating the angle between vectors when their dot product and magnitudes are known.
In the context of our example, we calculate the cosine of the angle between the normal vectors using the formula:

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

After solving, \( \cos \theta = \frac{1}{\sqrt{2}} \). To find \( \theta \), we apply the inverse cosine:

• \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \)

The result is \( \theta = 45^\circ \) or \( \frac{\pi}{4} \) radians. The inverse cosine is an essential function in trigonometry, allowing us to translate between angle measurements and scalar values from dot products.