Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(\mathbf{a}=\langle\sqrt{3} / 3, \sqrt{3} / 3, \sqrt{3} / 3\rangle, \mathbf{b}=\langle 1,-1,0\rangle\), and \(\mathbf{c}=\langle-2,-2,1\rangle\). Find the angle between each pair of vectors.

Step-by-Step Solution

Answer

Find angles: 1) \( \theta_{\mathbf{a},\mathbf{b}} = 90^{\circ} \), 2) \( \theta_{\mathbf{a},\mathbf{c}} \approx 125.26^{\circ} \), 3) \( \theta_{\mathbf{b},\mathbf{c}} = 90^{\circ} \).

Step 1: Understanding the Dot Product Formula

The dot product formula for two vectors \( \mathbf{x} = \langle x_1, x_2, x_3 \rangle \) and \( \mathbf{y} = \langle y_1, y_2, y_3 \rangle \) is given by \( \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 + x_3y_3 \). This formula helps in finding the cosine of the angle between the vectors.

Step 2: Calculating Dot Products

Calculate the dot products for each pair of vectors: 1. \( \mathbf{a} \cdot \mathbf{b} = \sqrt{3}/3 \cdot 1 + \sqrt{3}/3 \cdot (-1) + \sqrt{3}/3 \cdot 0 = 0 \).2. \( \mathbf{a} \cdot \mathbf{c} = \sqrt{3}/3 \cdot (-2) + \sqrt{3}/3 \cdot (-2) + \sqrt{3}/3 \cdot 1 = -3\sqrt{3}/3 = -\sqrt{3} \).3. \( \mathbf{b} \cdot \mathbf{c} = 1 \cdot (-2) + (-1) \cdot (-2) + 0 \cdot 1 = 0 \).

Step 3: Magnitudes of Vectors

Calculate the magnitudes of the vectors: 1. \( \|\mathbf{a}\| = \sqrt{(\sqrt{3}/3)^2 + (\sqrt{3}/3)^2 + (\sqrt{3}/3)^2} = \sqrt{3}/3 \times \sqrt{3} = 1 \).2. \( \|\mathbf{b}\| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2} \).3. \( \|\mathbf{c}\| = \sqrt{(-2)^2 + (-2)^2 + 1^2} = 3 \).

Step 4: Calculating Cosine of the Angles

Use the formula for the cosine of the angle between two vectors \( \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\| \|\mathbf{y}\|} \):1. \( \cos \theta_{\mathbf{a},\mathbf{b}} = \frac{0}{1 \times \sqrt{2}} = 0 \), which means the angle is \( \frac{\pi}{2} \) or 90 degrees.2. \( \cos \theta_{\mathbf{a},\mathbf{c}} = \frac{-\sqrt{3}}{1 \times 3} = -\frac{\sqrt{3}}{3} \), use inverse cosine to find the angle.3. \( \cos \theta_{\mathbf{b},\mathbf{c}} = \frac{0}{\sqrt{2} \times 3} = 0 \), meaning the angle is \( \frac{\pi}{2} \) or 90 degrees.

Step 5: Interpreting the Cosine Values

1. For \( \mathbf{a} \) and \( \mathbf{b} \), the angle is \( 90^{\circ} \) because \( \cos \theta = 0 \).2. For \( \mathbf{b} \) and \( \mathbf{c} \), the angle is also \( 90^{\circ} \) because \( \cos \theta = 0 \).3. For \( \mathbf{a} \) and \( \mathbf{c} \), calculate the inverse cosine of \(-\frac{\sqrt{3}}{3}\) to determine the angle.

Key Concepts

Dot Product

The dot product is a key operation in vector algebra, primarily used to find the angle between vectors. It involves multiplying corresponding components of two vectors and summing the results. Given vectors \( \mathbf{x} = \langle x_1, x_2, x_3 \rangle \) and \( \mathbf{y} = \langle y_1, y_2, y_3 \rangle \), their dot product is expressed as:

• \( \mathbf{x} \cdot \mathbf{y} = x_1y_1 + x_2y_2 + x_3y_3 \)

This formula is not only useful in finding the angle between vectors but also plays a significant role in physics and computer graphics.
When the dot product of two vectors equals zero, it's a clear indication that the vectors are perpendicular to each other.
In our exercise, the dot products for vector pairs \( \mathbf{a} \) and \( \mathbf{b} \), as well as \( \mathbf{b} \) and \( \mathbf{c} \), were both zero. This tells us these vector pairs are orthogonal or at a 90-degree angle from each other.

Magnitude of a Vector

The magnitude, or length, of a vector is a measure of how long the vector is. It is derived from the Pythagorean theorem and provides the scalar length of the vector. For any vector \( \mathbf{x} = \langle x_1, x_2, x_3 \rangle \), its magnitude is determined as:

• \( \|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2 + x_3^2} \)

This magnitude formula is essential when computing the angle between vectors using the dot product.
By finding the magnitudes of \( \mathbf{a} \), \( \mathbf{b} \), and \( \mathbf{c} \), we were able to proceed with finding the cosine of the angle between the vectors, which is crucial in vector analysis.
For instance, vector \( \mathbf{a} \) had a magnitude of 1, which means it is a unit vector, usually indicating a normalized direction with no scale.

Cosine of an Angle

The cosine of the angle between two vectors is derived from the dot product and magnitudes of the vectors, using the formula:

• \( \cos \theta = \frac{\mathbf{x} \cdot \mathbf{y}}{\|\mathbf{x}\| \|\mathbf{y}\|} \)

This ratio helps us understand not just the direction but the angular relationship between vectors. It gives insight into whether the vectors point in the same direction (cosine value close to 1), are perpendicular (cosine value zero), or point in opposite directions (cosine value close to -1).
In our example, the calculations revealed that \( \cos \theta_{\mathbf{a},\mathbf{b}} \) and \( \cos \theta_{\mathbf{b},\mathbf{c}} \) equal zero, showing those pairs of vectors are perpendicular.
Meanwhile, for \( \mathbf{a} \) and \( \mathbf{c} \), a cosine value of \(-\frac{\sqrt{3}}{3}\) indicates they form an acute angle, requiring the use of inverse cosine to find the exact angle in degrees or radians.