Problem 6 - 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. $$ \mathbf{u}=\mathbf{i}+\sqrt{3} \mathbf{j}, \quad \mathbf{v}=-\sqrt{3} \mathbf{i}+\mathbf{j} $$

Step-by-Step Solution

Answer

(a) \( \mathbf{u} \cdot \mathbf{v} = 0 \), (b) Angle = 90°.

Step 1: Calculate the Dot Product

The dot product of two vectors \( \mathbf{u} = (u_1, u_2) \) and \( \mathbf{v} = (v_1, v_2) \) is given by \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). First, identify the components of the vectors: \( \mathbf{u} = (1, \sqrt{3}) \) and \( \mathbf{v} = (-\sqrt{3}, 1) \). Calculate the dot product: \( \mathbf{u} \cdot \mathbf{v} = 1(-\sqrt{3}) + \sqrt{3}(1) = -\sqrt{3} + \sqrt{3} = 0 \).

Step 2: Magnitude of Vector \(\mathbf{u}\)

To find the magnitude of \( \mathbf{u} = (1, \sqrt{3}) \), use the formula \( ||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2} \). Calculating gives \( ||\mathbf{u}|| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2 \).

Step 3: Magnitude of Vector \(\mathbf{v}\)

To find the magnitude of \( \mathbf{v} = (-\sqrt{3}, 1) \), use the formula \( ||\mathbf{v}|| = \sqrt{v_1^2 + v_2^2} \). Calculating gives \( ||\mathbf{v}|| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2 \).

Step 4: Calculate the Angle Between \(\mathbf{u}\) and \(\mathbf{v}\)

The angle \( \theta \) between two vectors is found using the formula \( \cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| ||\mathbf{v}||} \). Since \( \mathbf{u} \cdot \mathbf{v} = 0 \), \( \cos(\theta) = \frac{0}{2 \times 2} = 0 \). Therefore, \( \theta = \cos^{-1}(0) = 90^\circ \).

Step 5: Round the Angle to the Nearest Degree

In this particular problem, the angle is already a whole number: \( 90^\circ \). So, there's no need for further rounding.

Key Concepts

Dot Product

The dot product is a fundamental concept in vector algebra that measures the extent of parallelism between two vectors. It's also known as the scalar product because the result is a single number rather than another vector. For two vectors \( \mathbf{u} = (u_1, u_2) \) and \( \mathbf{v} = (v_1, v_2) \), the dot product is calculated as:

• \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \)

If the dot product is zero, it signifies that the vectors are perpendicular to each other.
In our example, vectors \( \mathbf{u} = (1, \sqrt{3}) \) and \( \mathbf{v} = (-\sqrt{3}, 1) \) result in a dot product calculation of:

• \( \mathbf{u} \cdot \mathbf{v} = 1(-\sqrt{3}) + \sqrt{3}(1) = -\sqrt{3} + \sqrt{3} = 0 \)

This confirms the vectors are perpendicular.

Magnitude of a Vector

The magnitude of a vector, often called its length or norm, is a measure of how long the vector is. To calculate the magnitude, use the formula:

• For vector \( \mathbf{u} = (u_1, u_2) \), the magnitude \( ||\mathbf{u}|| \) is \( \sqrt{u_1^2 + u_2^2} \)
• For vector \( \mathbf{v} = (v_1, v_2) \), the magnitude \( ||\mathbf{v}|| \) is \( \sqrt{v_1^2 + v_2^2} \)

In our solution, both \( \mathbf{u} \) and \( \mathbf{v} \) have a magnitude of 2:

• \( ||\mathbf{u}|| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2 \)
• \( ||\mathbf{v}|| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{4} = 2 \)

A consistent magnitude means each vector maintains its length regardless of its direction.

Angle Between Vectors

The angle between two vectors can reveal much about their directional relationship.
This angle, \( \theta \), can be found using the formula:

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

With a dot product of 0, the vectors are at right angles (perpendicular), hence:

• \( \cos(\theta) = \frac{0}{2 \times 2} = 0 \)

The angle corresponding to \( \cos^{-1}(0) \) is \( 90^\circ \).
Recognizing perpendicular vectors can be particularly useful in geometry and physics as this perpendicularity implies no component of one vector is in the direction of the other.