Problem 65 - Precalculus with Calculus Previews

Question

Find the angle between the given pair of vectors. Round your answer to two decimal places. $$ \langle 1,4\rangle,\langle 2,-1\rangle $$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 101.31°.

Step 1: Understand the formula

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

Step 2: Calculate the dot product

The dot product \( \mathbf{a} \cdot \mathbf{b} \) for vectors \( \langle 1, 4 \rangle \) and \( \langle 2, -1 \rangle \) is calculated as follows: \[ 1 \cdot 2 + 4 \cdot (-1) = 2 - 4 = -2 \] The dot product is \( -2 \).

Step 3: Find vector magnitudes

The magnitude of vector \( \mathbf{a} = \langle 1, 4 \rangle \) is \[ \|\mathbf{a}\| = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17} \]. The magnitude of vector \( \mathbf{b} = \langle 2, -1 \rangle \) is \[ \|\mathbf{b}\| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \].

Step 4: Plug values into the formula

Use the formula from Step 1 to find \( \cos \theta \): \[ \cos \theta = \frac{-2}{\sqrt{17} \times \sqrt{5}} = \frac{-2}{\sqrt{85}} \].

Step 5: Calculate \( \theta \)

To find the angle \( \theta \), calculate \( \theta = \cos^{-1} \left( \frac{-2}{\sqrt{85}} \right) \). Using a calculator, \( \theta \approx 101.31^{\circ} \).

Key Concepts

Dot Product

The dot product is a fundamental operation when working with vectors. It's a way to multiply two vectors, which gives us a scalar result, instead of another vector. For two vectors, \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product \( \mathbf{a} \cdot \mathbf{b} \) is calculated as:\[a_1 \cdot b_1 + a_2 \cdot b_2\]This means you multiply the corresponding components of the vectors and then add those results together.
For example, with vectors \( \langle 1, 4 \rangle \) and \( \langle 2, -1 \rangle \), the dot product is:

• \( 1 \times 2 = 2 \)
• \( 4 \times (-1) = -4 \)

Adding these together gives \( 2 - 4 = -2 \). This scalar value of \(-2\) plays a crucial role in finding the angle between the vectors.

Vector Magnitude

The magnitude of a vector can be thought of as its "length" in geometric terms. It's a measure of how long the vector is when plotted on a coordinate plane.
The magnitude \( \|\mathbf{v}\| \) of a vector \( \mathbf{v} = \langle x, y \rangle \) is calculated using the Pythagorean theorem:\[\|\mathbf{v}\| = \sqrt{x^2 + y^2}\]This formula ensures we get a positive length for the vector. For example, for vector \( \langle 1, 4 \rangle \), the magnitude is:\[\|\mathbf{a}\| = \sqrt{1^2 + 4^2} = \sqrt{17}\]Similarly, for vector \( \langle 2, -1 \rangle \), it is:\[\|\mathbf{b}\| = \sqrt{2^2 + (-1)^2} = \sqrt{5}\]Knowing the magnitudes helps us in determining the angle between two vectors using the dot product.

Inverse Cosine

Inverse Cosine, often denoted as \( \cos^{-1} \), is the operation used to find an angle when you know the cosine value of that angle.
In the context of vectors, once you have the ratio from the dot product and the magnitudes:\[\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\]you use the inverse cosine to find \( \theta \).

• This angle \( \theta \) will always be between 0 and 180 degrees since cosine is positive in the first quadrant and negative in the second.
• Simply apply \( \cos^{-1} \) to the ratio \( \frac{-2}{\sqrt{85}} \) found earlier.
• With a calculator, you get \( \theta \approx 101.31^{\circ} \).

This means the two vectors form an obtuse angle.