Problem 11 - Precalculus Mathematics for Calculus

Question

Find \((a) u \cdot v\) and \((b)\) the angle between \(u\) and \(v\) to the nearest degree. $$\mathbf{u}=-5 \mathbf{j}, \quad \mathbf{v}=-\mathbf{i}-\sqrt{3} \mathbf{j}$$

Step-by-Step Solution

Answer

(a) The dot product is \(5\sqrt{3}\). (b) The angle is \(30^{\circ}\).

Step 1: Understand the Dot Product Formula

The dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is given by \( u \cdot v = u_1v_1 + u_2v_2 \), where \( u_1, u_2 \) are the components of \( \mathbf{u} \) and \( v_1, v_2 \) are the components of \( \mathbf{v} \). Here, \( \mathbf{u} = 0\mathbf{i} - 5\mathbf{j}, \mathbf{v} = -1\mathbf{i} - \sqrt{3}\mathbf{j} \).

Step 2: Apply the Dot Product Formula

Substitute the components into the dot product formula: \( u \cdot v = (0)(-1) + (-5)(-\sqrt{3}) \). Calculate this to find the product.

Step 3: Calculate the Dot Product

Since \( u \cdot v = 0 + 5\sqrt{3} = 5\sqrt{3} \), we find that the dot product is \( 5\sqrt{3} \).

Step 4: Use the Formula for Angle Between Vectors

The angle \( \theta \) between two vectors can be found using \( \cos \theta = \frac{u \cdot v}{\|u\|\|v\|} \), where \( \|u\| \) and \( \|v\| \) are the magnitudes of \( \mathbf{u} \) and \( \mathbf{v} \).

Step 5: Find the Magnitudes of u and v

Calculate the magnitudes using \( \|u\| = \sqrt{0^2 + (-5)^2} = 5 \) and \( \|v\| = \sqrt{(-1)^2 + (\sqrt{3})^2} = 2 \).

Step 6: Substitute into the Angle Formula

Substitute the dot product and the magnitudes into the angle formula: \( \cos \theta = \frac{5\sqrt{3}}{5 \times 2} = \frac{\sqrt{3}}{2} \).

Step 7: Calculate the Angle

The angle corresponding to \( \cos \theta = \frac{\sqrt{3}}{2} \) is \( \theta = 30^{\circ} \). Thus, the angle between \( \mathbf{u} \) and \( \mathbf{v} \) to the nearest degree is \( 30^{\circ} \).

Key Concepts

Understanding the Dot Product

The dot product is a key concept in vector mathematics. It's an operation that takes two equal-length sequences of numbers (usually vectors) and returns a single number, known as a scalar. To compute the dot product of two vectors, \( \mathbf{u} \) and \( \mathbf{v} \), you use the formula:

• \( u \cdot v = u_1v_1 + u_2v_2 \)

Here, \( u_1 \) and \( v_1 \) are the first components of vectors \( \mathbf{u} \) and \( \mathbf{v} \), while \( u_2 \) and \( v_2 \) are their second components.

In the exercise, we have \( \mathbf{u} = 0 \mathbf{i} - 5 \mathbf{j} \) and \( \mathbf{v} = -1 \mathbf{i} - \sqrt{3} \mathbf{j} \). By substituting these into the dot product formula, we get:

• \( u \cdot v = (0)(-1) + (-5)(-\sqrt{3}) \)

This simplifies to \( 0 + 5\sqrt{3} = 5\sqrt{3} \). So, the dot product in this case is \( 5\sqrt{3} \).

Understanding the dot product helps in solving many problems involving angles and projections. It simplifies calculations and reveals how closely two vectors align.

Calculating the Angle Between Vectors

Determining the angle between vectors involves using the dot product. The formula to find the angle, \( \theta \), between vectors \( \mathbf{u} \) and \( \mathbf{v} \) is:

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

Where \( \|u\| \) and \( \|v\| \) are the magnitudes of the vectors. This equation relates the dot product to the cosine of the angle between the vectors, showing their directional relationship.

We've found \( u \cdot v = 5\sqrt{3} \). To find \( \theta \), we need \( \|u\| \) and \( \|v\| \). With \( \|u\| = 5 \) and \( \|v\| = 2 \), we substitute into the formula:

• \( \cos \theta = \frac{5\sqrt{3}}{5 \times 2} = \frac{\sqrt{3}}{2} \)

The angle \( \theta \) is thus the angle whose cosine is \( \frac{\sqrt{3}}{2} \), which is \( 30^{\circ} \).

Finding this angle reveals whether the vectors are approaching orthogonality or parallelism.

Understanding Vector Magnitude

To understand vectors, it's crucial to grasp vector magnitude. A vector's magnitude is its length or size, a measure of how "strong" the vector is. The formula to calculate the magnitude of a vector \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} \) is:

• \( \|a\| = \sqrt{a_1^2 + a_2^2} \)

This equation echoes the Pythagorean theorem applied in the context of space—a vector is often visualized as an arrow with a specific direction and length.

For vector \( \mathbf{u} = 0 \mathbf{i} - 5 \mathbf{j} \), its magnitude is:

• \( \|u\| = \sqrt{0^2 + (-5)^2} = 5 \)

Similarly, for \( \mathbf{v} = -1 \mathbf{i} - \sqrt{3} \mathbf{j} \):

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

These magnitudes show how long each vector is, without regard to their direction.

Understanding vector magnitude is essential for real-world applications like force, velocity, and other measurements, providing insight into the "strength" of vector quantities.