Problem 15 - Precalculus

Question

Find the angle (round to the nearest degree) between each pair of vectors. $$\langle-2,-3\rangle \text { and }\langle-3,4\rangle$$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 122 degrees.

Step 1: Understand the formula

The formula for the angle \( \theta \) between two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) is given by \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \). Here, \( \mathbf{a} \cdot \mathbf{b} \) is the dot product and \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \) are the magnitudes of the vectors.

Step 2: Calculate the dot product

Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \) for vectors \( \langle -2, -3 \rangle \) and \( \langle -3, 4 \rangle \).\[ \mathbf{a} \cdot \mathbf{b} = (-2)(-3) + (-3)(4) = 6 - 12 = -6 \]

Step 3: Find the magnitudes of the vectors

Calculate the magnitudes of the vectors \( \langle -2, -3 \rangle \) and \( \langle -3, 4 \rangle \).\[ \|\mathbf{a}\| = \sqrt{(-2)^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \]\[ \|\mathbf{b}\| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Step 4: Calculate the cosine of the angle

Substitute the dot product and magnitudes into the cosine formula.\[ \cos \theta = \frac{-6}{\sqrt{13} \times 5} = \frac{-6}{5\sqrt{13}} \]

Step 5: Compute the angle \( \theta \)

Find the angle by taking the arccosine of \( \frac{-6}{5\sqrt{13}} \).\[ \theta = \cos^{-1}\left(\frac{-6}{5\sqrt{13}}\right) \approx 122^\circ \]

Key Concepts

Dot Product

When dealing with vectors, the dot product is a fundamental operation that comes in handy for various calculations, such as finding the angle between vectors. It is a way to multiply two vectors to get a scalar (a single number instead of another vector). Here's how it works:

The dot product for two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) is calculated as:

• Multiply the corresponding components of the vectors: \( a_1 \times b_1 \) and \( a_2 \times b_2 \).
• Add the results: \( a_1 \times b_1 + a_2 \times b_2 \).

This gives you a value that tells you about the vectors' alignment. A positive dot product implies the vectors point roughly in the same direction, while a negative value indicates they point in opposite directions. If the dot product is zero, the vectors are perpendicular.

In our case, the dot product for vectors \( \langle -2, -3 \rangle \) and \( \langle -3, 4 \rangle \) turned out to be \(-6\), which means they are not aligned in the same direction, giving us a hint about the angle between them.

Magnitude of Vectors

The magnitude of a vector involves determining its length, much like finding the length of a line segment. It is calculated using the components of the vector through the Pythagorean Theorem. If you have a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), its magnitude \( \|\mathbf{a}\| \) is calculated as follows:

• Square each component of the vector.
• Add the squared values.
• Take the square root of the sum.

For vector \( \langle -2, -3 \rangle \), the magnitude is \( \sqrt{13} \), and for vector \( \langle -3, 4 \rangle \), it is \( 5 \). These magnitudes reflect the lengths of the vectors. They are essential in our calculation because they are part of the structure when determining the angle between vectors. Specifically, they help normalize the vectors to find the \( \cos \theta \), which is crucial for finding the angle.

Cosine of Angle

The cosine of the angle between two vectors is directly related to the dot product and the magnitudes of the vectors. This is critical in vector mathematics for determining the exact measure of the angle formed between two vectors.

The formula for the cosine of the angle \( \theta \) between vectors \( \mathbf{a} \) and \( \mathbf{b} \) is:

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

This formula essentially normalizes the dot product by dividing it by the product of the magnitudes of the vectors. It yields a value between -1 and 1:

• If \( \cos \theta = 1 \), the vectors are perfectly aligned.
• If \( \cos \theta = 0 \), the vectors are perpendicular.
• If \( \cos \theta = -1 \), the vectors are opposed 180 degrees apart.

In the example with vectors \( \langle -2, -3 \rangle \) and \( \langle -3, 4 \rangle \), we computed \( \cos \theta \) as \( \frac{-6}{5\sqrt{13}} \), which results in an angle of approximately \( 122^\circ \). This means the vectors form an obtuse angle, indicating they point in very different directions.