Problem 11 - 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. $$ \mathbf{u}=-5 \mathbf{j}, \quad \mathbf{v}=-\mathbf{i}-\sqrt{3} \mathbf{j} $$

Step-by-Step Solution

Answer

(a) \( \mathbf{u} \cdot \mathbf{v} = 5\sqrt{3} \) and (b) the angle is approximately 30 degrees.

Step 1: Identify Components of the Vectors

The components of the vector \( \mathbf{u} \) are \( (0, -5) \) and the components of the vector \( \mathbf{v} \) are \( (-1, -\sqrt{3}) \). These are obtained from the given vectors \( \mathbf{u} = -5\mathbf{j} \) and \( \mathbf{v} = -\mathbf{i} - \sqrt{3}\mathbf{j} \).

Step 2: Calculate the Dot Product \( \mathbf{u} \cdot \mathbf{v} \)

The dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated by multiplying the corresponding components and summing them: \( \mathbf{u} \cdot \mathbf{v} = (0)(-1) + (-5)(-\sqrt{3}) = 0 + 5\sqrt{3} = 5\sqrt{3} \).

Step 3: Calculate the Magnitude of Each Vector

The magnitude of \( \mathbf{u} \) is \( \| \mathbf{u} \| = \sqrt{0^2 + (-5)^2} = 5 \). The magnitude of \( \mathbf{v} \) is \( \| \mathbf{v} \| = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = 2 \).

Step 4: Use the Dot Product to Find the Cosine of the Angle

The formula to find the cosine of the angle \( \theta \) between two vectors is \( \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\| \mathbf{u} \| \| \mathbf{v} \|} \). Substitute the values from previous steps: \( \cos\theta = \frac{5\sqrt{3}}{5 \times 2} = \frac{\sqrt{3}}{2} \).

Step 5: Calculate the Angle \( \theta \) in Degrees

To find the angle \( \theta \) (in degrees), use the inverse cosine function: \( \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \). This results in \( \theta \approx 30 ^{\circ} \).

Key Concepts

Angle Between Vectors

When dealing with vectors, finding the angle between them is a valuable skill. It helps understand their direction in relation to each other. To find the angle between two vectors, say \( \mathbf{u} \) and \( \mathbf{v} \), you first need their dot product and their magnitudes. Using this information, the formula to determine the angle \( \theta \) is derived from the relationship:

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

This formula relates the dot product directly to cosine, which is why it can give the angle between the vectors once you know \( \cos(\theta) \). Once you compute the cosine of the angle, using the inverse cosine function will yield the angle \( \theta \). For the exercise given, it was calculated as approximately \( 30^\circ \). Understanding these calculations provides insight into the spatial relationship between \( \mathbf{u} \) and \( \mathbf{v} \).
Using vectors in such manner is a fundamental aspect of linear algebra and physics.

Magnitude of a Vector

The magnitude of a vector is like the length of the vector. It's how much distance the point moves from the origin to its terminal location in space. If you look at a vector \( \mathbf{u} = a \mathbf{i} + b \mathbf{j} \), the magnitude \( \| \mathbf{u} \| \) is calculated with the Pythagorean theorem:

• \( \| \mathbf{u} \| = \sqrt{a^2 + b^2} \)

This comes from thinking of \( \mathbf{u} \) as the hypotenuse of a right triangle with legs of length \( a \) and \( b \). In the given exercise, the vector \( \mathbf{u} = -5\mathbf{j} \) has a magnitude of 5 because it only lies along the \( y \)-axis, and thus \( \| \mathbf{u} \| = \sqrt{0^2 + (-5)^2} = 5 \). Similarly, for the vector \( \mathbf{v} = -\mathbf{i} - \sqrt{3}\mathbf{j} \), the magnitude calculated is 2. Understanding the concept of magnitude helps determine how much impact a vector has in physics or its overall spatial extent in other applications.

Cosine of the Angle

The cosine of the angle between two vectors is a crucial element in vector analysis. It provides not only the measure of the angle but also indicates how much two vectors overlap in their direction. Mathematically, with two vectors \( \mathbf{u} \) and \( \mathbf{v} \), the cosine of the angle \( \theta \) between them is determined by:

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

Here, the dot product represents the extent of one vector's projection onto the other, and the product of their magnitudes normalizes this length to make it a proper cosine angle measure. In the exercise's context, it resulted as \( \frac{\sqrt{3}}{2} \), meaning a fairly significant directional alignment, indicative of an angle close to 30 degrees. Calculating \( \cos(\theta) \) is fundamental, especially in determining how vectors are oriented in physics, engineering, and computer graphics.