Problem 67 - Precalculus with Calculus Previews

Question

Find the angle between the given pair of vectors. Round your answer to two decimal places. $$ \mathbf{i}-\mathbf{j}, 3 \mathbf{i}+\mathbf{j} $$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 63.43°.

Step 1: Identify the given vectors

The two vectors provided are \( \mathbf{a} = \mathbf{i} - \mathbf{j} \) and \( \mathbf{b} = 3\mathbf{i} + \mathbf{j} \). Express them in the standard form as \( \mathbf{a} = (1, -1) \) and \( \mathbf{b} = (3, 1) \).

Step 2: Use the dot product formula

The formula for the dot product of two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \) is \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \). Substituting the values of \( \mathbf{a} \) and \( \mathbf{b} \), we compute: \( \mathbf{a} \cdot \mathbf{b} = 1 \cdot 3 + (-1) \cdot 1 = 3 - 1 = 2 \).

Step 3: Calculate the magnitude of each vector

The magnitude of vector \( \mathbf{a} = (1, -1) \) is \( |\mathbf{a}| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \). The magnitude of vector \( \mathbf{b} = (3, 1) \) is \( |\mathbf{b}| = \sqrt{3^2 + 1^2} = \sqrt{10} \).

Step 4: Apply the angle formula between vectors

The angle \( \theta \) between two vectors can be found using the formula \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \cdot |\mathbf{b}|} \). Substituting the known values: \( \cos \theta = \frac{2}{\sqrt{2} \cdot \sqrt{10}} = \frac{2}{\sqrt{20}} = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}} \).

Step 5: Calculate the angle using inverse cosine

To find \( \theta \), we take the inverse cosine (arccos) of \( \frac{1}{\sqrt{5}} \). Compute \( \theta = \arccos \left( \frac{1}{\sqrt{5}} \right) \). Using a calculator, this gives \( \theta \approx 63.43^\circ \).

Step 6: Round the angle to two decimal places

The calculated angle \( \theta \) is approximately \( 63.43^\circ \), already rounded to two decimal places.

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a way of combining two vectors to result in a single number, or scalar. It provides insights into how much of one vector goes in the direction of another. To find the dot product of two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), use the formula:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \]In our exercise, vectors \( \mathbf{a} = (1, -1) \) and \( \mathbf{b} = (3, 1) \) give a dot product of:

• \( 1 \times 3 + (-1) \times 1 = 3 - 1 = 2 \)

The dot product can be thought of as a projection, where you project one vector onto another. If the dot product is positive, as it is here, the vectors are pointing somewhat in the same general direction.

Magnitude of a Vector

The magnitude of a vector, often referred to as its length or norm, tells us how long the vector is. Imagine you're measuring the length of an arrow in space—the magnitude is this measure. For a vector \( \mathbf{a} = (a_1, a_2) \), the magnitude \( |\mathbf{a}| \) is found using the Pythagorean theorem:\[ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2} \]For \( \mathbf{a} = (1, -1) \):

• \( |\mathbf{a}| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \)

For \( \mathbf{b} = (3, 1) \):

• \( |\mathbf{b}| = \sqrt{3^2 + 1^2} = \sqrt{9 + 1} = \sqrt{10} \)

Understanding vector magnitude is crucial in applications ranging from physics to computer graphics, where it helps in normalizing vectors and calculating distances.

Inverse Cosine

The inverse cosine function, denoted as \( \cos^{-1} \) or \( \arccos \), helps us find the angle between two vectors when we know their dot product and magnitudes. When vectors \( \mathbf{a} \) and \( \mathbf{b} \) have a known dot product \( \mathbf{a} \cdot \mathbf{b} \) and magnitudes \( |\mathbf{a}| \) and \( |\mathbf{b}| \), the angle \( \theta \) is found using the formula:\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \cdot |\mathbf{b}|} \]Thus, to find \( \theta \), we need to use the inverse cosine function:

• \( \theta = \cos^{-1} \left( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \cdot |\mathbf{b}|} \right) \)
• Here, \( \theta = \cos^{-1} \left( \frac{2}{\sqrt{2} \times \sqrt{10}} \right) \approx 63.43^\circ \)

Using a calculator is the simplest way to compute this, and it helps translate the abstract concept into a concrete angle value, showing how these vectors orient in space relative to each other.