Problem 14 - Algebra and Trigonometry

Question

Find (a) \(\mathbf{u} \cdot \mathbf{v}\) and (b) the angle between \(\mathbf{u}\) and \(\mathbf{v}\) to the nearest degree. $$ \mathbf{u}=3 \mathbf{i}+4 \mathbf{j}, \quad \mathbf{v}=-2 \mathbf{i}-\mathbf{j} $$

Step-by-Step Solution

Answer

(a) -10, (b) 153°

Step 1: Calculate Dot Product

To find the dot product \( \mathbf{u} \cdot \mathbf{v} \), use the formula \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). Here, \( u_1 = 3 \), \( u_2 = 4 \), \( v_1 = -2 \), and \( v_2 = -1 \). Thus, the calculation is:\[\mathbf{u} \cdot \mathbf{v} = (3)(-2) + (4)(-1) = -6 - 4 = -10.\]

Step 2: Compute Magnitudes

Find the magnitude of \( \mathbf{u} \) and \( \mathbf{v} \). The magnitude of \( \mathbf{u} \) is\[\|\mathbf{u}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5.\]The magnitude of \( \mathbf{v} \) is\[\|\mathbf{v}\| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}.\]

Step 3: Calculate Cosine of Angle

Use the formula for the cosine of the angle \( \theta \) between two vectors: \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \). Substituting the known values:\[\cos \theta = \frac{-10}{5 \times \sqrt{5}} = \frac{-10}{5\sqrt{5}} = \frac{-2}{\sqrt{5}}.\]

Step 4: Find the Angle in Degrees

To find \( \theta \), compute \( \theta = \cos^{-1}\left( \frac{-2}{\sqrt{5}} \right) \). Use a calculator to find the angle \( \theta \approx 153.43^{\circ} \), which rounds to the nearest degree as \( 153^{\circ} \).

Key Concepts

Vector Magnitudes

When dealing with vectors, understanding magnitudes is crucial. A vector, like \( \mathbf{u} \) and \( \mathbf{v} \) in this problem, has both direction and magnitude. The magnitude can be thought of as the length or size of the vector. It is denoted by \( \|\mathbf{u}\| \) for vector \( \mathbf{u} \).To calculate the magnitude of a vector like \( \mathbf{u} = 3\mathbf{i} + 4\mathbf{j} \), use the formula:

• \( \|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2} \).

For \( \mathbf{u} \), this gives \( \|\mathbf{u}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \). This tells us the length of the vector is 5 units.
Similarly, compute \( \|\mathbf{v}\| \) for \( \mathbf{v} = -2\mathbf{i} - \mathbf{j} \):

• \( \|\mathbf{v}\| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \).

This gives a magnitude of approximately 2.24 units. Remember, the magnitude is always positive, reflecting the distance aspect of the vector.

Cosine of Angle

The cosine of the angle between two vectors is a measure of the direction of one vector relative to another. This is a key part of determining how similar two vectors are in terms of direction.To find the cosine of the angle \( \theta \) between vectors \( \mathbf{u} \) and \( \mathbf{v} \), use the dot product \( \mathbf{u} \cdot \mathbf{v} \) and their magnitudes:

• \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \).

In this exercise, both \( \mathbf{u} \) and \( \mathbf{v} \) have been calculated designed to use this formula efficiently.
The result of the dot product is \(-10\), while the magnitudes are \(5\) and \(\sqrt{5}\), leading to:

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

This value represents how much the direction of the two vectors point in opposite directions, with negative results indicating opposite directional trends.

Angle Between Vectors

The angle between vectors helps us understand the geometric relationship between them. To find the angle \( \theta \) itself, we first find the cosine of the angle using vector measures and then apply the inverse cosine function to extract \( \theta \).In mathematical language, use the formula:

• \( \theta = \cos^{-1}\left( \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \right) \).

For our vectors \( \mathbf{u} \) and \( \mathbf{v} \), we calculated the cosine value as \( \frac{-2}{\sqrt{5}} \). To find the actual angle, compute:

• \( \theta = \cos^{-1}\left( \frac{-2}{\sqrt{5}} \right) \).

A calculator provides \( \theta \approx 153.43^{\circ} \). This angle suggests a significant directional difference, as it is more than 90 degrees, implying that the vectors are pointed more in opposite than in similar directions. Rounding this gives a succinct result: \( 153^{\circ} \).