Problem 13 - 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. $$ \mathbf{u}=\mathbf{i}+3 \mathbf{j}, \quad \mathbf{v}=4 \mathbf{i}-\mathbf{j} $$

Step-by-Step Solution

Answer

(a) \(1\), (b) \(87^\circ\)

Step 1: Calculate the dot product

The dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated using the formula: \( u_1v_1 + u_2v_2 \). For \( \mathbf{u} = \mathbf{i} + 3\mathbf{j} \) and \( \mathbf{v} = 4\mathbf{i} - \mathbf{j} \), this becomes: \( 1 \cdot 4 + 3 \cdot (-1) = 4 - 3 = 1 \). Thus, \( \mathbf{u} \cdot \mathbf{v} = 1 \).

Step 2: Calculate the magnitudes of the vectors

The magnitude of a vector \( \mathbf{u} = a\mathbf{i} + b\mathbf{j} \) is given by \( \sqrt{a^2 + b^2} \). For \( \mathbf{u} = \mathbf{i} + 3\mathbf{j} \), the magnitude is \( \sqrt{1^2 + 3^2} = \sqrt{10} \). For \( \mathbf{v} = 4\mathbf{i} - \mathbf{j} \), the magnitude is \( \sqrt{4^2 + (-1)^2} = \sqrt{17} \).

Step 3: Use the dot product to find the angle

The angle \( \theta \) between two vectors can be found using the equation: \( \cos^{-1}\left(\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\right) \). Substitute the values: \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{10} \cdot \sqrt{17}}\right) \approx \cos^{-1}(0.0767) \).

Step 4: Calculate the angle in degrees

Calculate \( \theta \) using the inverse cosine function on a calculator: \( \theta \approx \cos^{-1}(0.0767) \approx 86.5^\circ \). Round this to the nearest degree to get \( \theta \approx 87^\circ \).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector calculus. It helps us determine the interaction between two vectors by providing a scalar result. To compute the dot product of two vectors \( \mathbf{u} = \mathbf{i} + 3\mathbf{j} \) and \( \mathbf{v} = 4\mathbf{i} - \mathbf{j} \), we follow the formula:

• Multiply the corresponding components: \( u_1v_1 + u_2v_2 \).
• Here, \( u_1 = 1, u_2 = 3 \) and \( v_1 = 4, v_2 = -1 \).
• Calculate: \( 1 \cdot 4 + 3 \cdot (-1) = 4 - 3 = 1 \).

The dot product \( \mathbf{u} \cdot \mathbf{v} \) is then 1. This simple operation gives us insight into how aligned the vectors are. A positive dot product indicates that vectors are pointing in generally the same direction, while a negative result shows they are more opposite.

Vector Magnitude

Understanding the magnitude of a vector helps measure its length or size. It's like determining the "distance" a vector covers from its tail to its tip. Let's break it down for each vector:

• For vector \( \mathbf{u} = \mathbf{i} + 3\mathbf{j} \), the magnitude \( \|\mathbf{u}\| \) is found using \( \sqrt{a^2 + b^2} \), where \( a = 1 \) and \( b = 3 \).
• This yields \( \sqrt{1^2 + 3^2} = \sqrt{10} \).
• For vector \( \mathbf{v} = 4\mathbf{i} - \mathbf{j} \), we use \( a = 4 \) and \( b = -1 \).
• The magnitude \( \|\mathbf{v}\| \) then becomes \( \sqrt{4^2 + (-1)^2} = \sqrt{17} \).

These magnitudes describe how "long" each vector is, which is crucial for calculating angles between them and understanding their physical representations in space.

Angle Between Vectors

Finding the angle between vectors allows us to understand their spatial relationship. The angle \( \theta \) provides a measure of how much one vector "turns" to align with another. We use this formula: \[ \cos^{-1}\left(\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\right) \] Doing the math with our vectors:

• The dot product \( \mathbf{u} \cdot \mathbf{v} = 1 \).
• The magnitudes are \( \sqrt{10} \) and \( \sqrt{17} \).
• Plug these into the formula: \( \theta = \cos^{-1}\left(\frac{1}{\sqrt{10} \cdot \sqrt{17}}\right) \approx \cos^{-1}(0.0767) \).

Using a calculator, \( \theta \approx 86.5^\circ \). Rounding this value gives us \( 87^\circ \). This angle tells us that vectors \( \mathbf{u} \) and \( \mathbf{v} \) are almost perpendicular, as 90 degrees signifies a perpendicular relationship in vector terms.