Problem 67 - A Graphical Approach to Precalculus with Limits

Question

Find the angle between each pair of vectors. $$\langle 2,1\rangle,\langle- 3,1\rangle$$

Step-by-Step Solution

Answer

The angle between the vectors is \(135^\circ\) or \(\frac{3\pi}{4}\) radians.

Step 1: Understanding the Formula for Angle Between Two Vectors

To find the angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \), we use the formula: \[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \cdot \|\mathbf{b}\|}\]Where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product of \( \mathbf{a} \) and \( \mathbf{b} \) and \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \) are the magnitudes of the vectors.

Step 1: Calculate the Dot Product

The dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as follows: \[\mathbf{a} \cdot \mathbf{b} = a_1 \cdot b_1 + a_2 \cdot b_2\]For the vectors \( \mathbf{a} = \langle 2, 1 \rangle \) and \( \mathbf{b} = \langle -3, 1 \rangle \): \[\mathbf{a} \cdot \mathbf{b} = (2)(-3) + (1)(1) = -6 + 1 = -5\]

Step 2: Calculate the Magnitude of Each Vector

The magnitude of a vector \( \mathbf{a} \) is \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \).For \( \mathbf{a} = \langle 2, 1 \rangle \): \[\|\mathbf{a}\| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}\]For \( \mathbf{b} = \langle -3, 1 \rangle \): \[\|\mathbf{b}\| = \sqrt{(-3)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10}\]

Step 3: Apply the Formula to Find Cosine of the Angle

Using the values we've calculated, substitute them into the formula for cosine of the angle:\[\cos \theta = \frac{-5}{\sqrt{5} \times \sqrt{10}} \]Simplify the expression:\[\cos \theta = \frac{-5}{\sqrt{50}} = \frac{-5}{\sqrt{25 \times 2}} = \frac{-5}{5\sqrt{2}} = \frac{-1}{\sqrt{2}}\]

Step 4: Calculate the Angle

Using the value for \( \cos \theta \), calculate the inverse cosine to find \( \theta \):\[\theta = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)\]This is equivalent to \( \theta = 135^\circ \) or \( \theta = \frac{3\pi}{4} \) radians. You can use a calculator to find that \( \theta \approx 135^\circ \).

Key Concepts

Dot Product

The dot product is an essential operation when working with vectors. It simplifies the interaction between vectors by mapping them onto a single number, called a scalar. This computation helps us establish relationships between vectors in terms of angles.
To find the dot product of vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), we use the formula:

• \( \mathbf{a} \cdot \mathbf{b} = a_1 \cdot b_1 + a_2 \cdot b_2 \)

Here’s an example with vectors: For \( \mathbf{a} = \langle 2,1 \rangle \) and \( \mathbf{b} = \langle -3,1 \rangle \), calculate:

• \( \mathbf{a} \cdot \mathbf{b} = (2)(-3) + (1)(1) = -6 + 1 = -5 \)

The outcome \(-5\) shows how \( \mathbf{a} \) and \( \mathbf{b} \) relate spatially. A positive dot product would indicate vectors pointing in the same general direction, zero for perpendicular vectors, and negative for vectors pointing in opposite directions.

Vector Magnitude

The magnitude of a vector is a measure of its length or size. Understanding vector magnitude is crucial because it provides additional insight into the vector's representation without consideration of its direction.
For any given vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), the magnitude \( \|\mathbf{a}\| \) is calculated using:

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

Let's compute the magnitude for our example vectors:

• For \( \mathbf{a} = \langle 2,1 \rangle \), \( \|\mathbf{a}\| = \sqrt{2^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \)
• For \( \mathbf{b} = \langle -3,1 \rangle \), \( \|\mathbf{b}\| = \sqrt{(-3)^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \)

Vector magnitude acts like a ruler that quantifies the vector's reach in space. This is integral when comparing vectors or working with operations like the dot product and angle calculations.

Trigonometric Functions

Trigonometric functions help bridge the relationship between angles and side lengths in right-angled triangles. In vector mathematics, they are essential for determining angles between vectors.
The key formula used in vector math for finding angles is:

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

This equation shows how the dot product and magnitudes of vectors \( \mathbf{a} \) and \( \mathbf{b} \) are related to the cosine of the angle \( \theta \) between them. In the example provided, substituting the values, we get:

• \( \cos \theta = \frac{-5}{\sqrt{5} \times \sqrt{10}} = \frac{-1}{\sqrt{2}} \)

By finding \( \theta \) using inverse cosine, we convert the cosine value back to an angle:

• \( \theta = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) \)
Ultimately, this calculation helps us understand how vectors position and orient relative to each other in space. Applying trigonometric functions ensures precise angle assessment, essential for fields like physics and engineering.