Problem 92 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use the Law of cosines to find the angle \(\alpha\) between the vectors. (Assume \(0^{\circ} \leq \alpha \leq 180^{\circ}\) ). $$\mathbf{v}=3 \mathbf{i}+\mathbf{j}, \quad \mathbf{w}=2 \mathbf{i}-\mathbf{j}$$

Step-by-Step Solution

Answer

The angle \(\alpha\) between the vectors \(\mathbf{v}\) and \(\mathbf{w}\) is approximately \(45\) degrees.

Step 1: Find the magnitudes of the vectors

The magnitude of a vector is calculated as the square root of the sum of the squares of its components. For vector \(\mathbf{v}\), the magnitude will be \(||\mathbf{v}|| = \sqrt{(3^2 + 1^2)}= \sqrt{10}\). Similarly, the magnitude of vector \(\mathbf{w}\) is \(||\mathbf{w}|| = \sqrt{(2^2 + (-1)^2)}= \sqrt{5}\).

Step 2: Calculate the dot product of the vectors

The dot product of two vectors is the sum of the products of their corresponding components. In this case, it will be \(\mathbf{v} \cdot \mathbf{w} = (3*2 + 1*(-1)) = 5.\)

Step 3: Apply the Law of Cosines

The Law of Cosines is used to find the angle between two vectors and is given by \(\cos{\alpha} = \frac{\mathbf{v} \cdot \mathbf{w}}{||\mathbf{v}||*||\mathbf{w}||}\). Substituting the obtained values, we have \(\cos{\alpha} = \frac{5}{\sqrt{10} * \sqrt{5}} = \frac{5}{\sqrt{50}} = \frac{\sqrt{2}}{2}\). Then, by using the inverse cosine function, the angle \(\alpha\) in radians is \(\alpha = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right)\). To express the angle in degrees, use the conversion \(1 \text{ radian} = \frac{180}{\pi} \text{ degrees}\), which gives \(\alpha = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) * \frac{180}{\pi} \approx 45^\circ\).

Key Concepts

Dot Product

The dot product is a way to multiply two vectors, resulting in a scalar value. It's calculated by multiplying corresponding components of each vector and then summing those products.
For two vectors \( \mathbf{a} = a_1 \mathbf{i} + a_2 \mathbf{j} \) and \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} \), their dot product \( \mathbf{a} \cdot \mathbf{b} \) is given by: \[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 \] The dot product provides a measure of how much one vector goes in the direction of the other.
If the dot product is zero, the vectors are perpendicular. In our given vectors \( \mathbf{v} = 3 \mathbf{i} + \mathbf{j} \) and \( \mathbf{w} = 2 \mathbf{i} - \mathbf{j} \), the dot product was calculated as: \( 3 \cdot 2 + 1 \cdot (-1) = 5 \).
This tells us there's some alignment between \( \mathbf{v} \) and \( \mathbf{w} \). Understanding dot products is crucial for finding angles between vectors.

Vector Magnitudes

A vector's magnitude is essentially its length, represented as the distance from the tail to the tip in geometric terms. For a 2D vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} \), the magnitude is calculated using the formula: \[ ||\mathbf{v}|| = \sqrt{a^2 + b^2} \] This formula is derived from the Pythagorean theorem.
Knowing the vector magnitudes of \( \mathbf{v} = 3 \mathbf{i} + \mathbf{j} \) and \( \mathbf{w} = 2 \mathbf{i} - \mathbf{j} \), we found: \( ||\mathbf{v}|| = \sqrt{10} \) and \( ||\mathbf{w}|| = \sqrt{5} \). Understanding vector magnitudes helps in determining the relative sizes of vectors, which is useful when calculating angles between them.
Magnitudes play a key role when combined with the dot product to find angles using the Law of Cosines.

Inverse Cosine Function

The inverse cosine function, denoted as \( \cos^{-1} \), is used to find the angle whose cosine is a given value. In the context of vectors, after obtaining the cosine of the angle from the dot product and magnitudes, we utilize \( \cos^{-1} \) to determine the actual angle between the vectors.
The range of the inverse cosine function is from 0° to 180°, which is perfect for calculating angles between vectors, as they can be reasonably big or small. For our example, solving the equation: \( \cos{\alpha} = \frac{\sqrt{2}}{2} \) led to: \( \alpha = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) \).
This result, in degrees, gives us approximately 45°, indicating the vectors are neither perpendicular nor parallel, but at an intermediate angle. Understanding the inverse cosine function is fundamental in translating between trigonometric ratios and angle measures.