Problem 73 - A Graphical Approach to Precalculus with Limits

Question

Find the angle between each pair of vectors. $$\mathbf{i}+\mathbf{j}, 3 \mathbf{i}+4 \mathbf{j}$$

Step-by-Step Solution

Answer

The angle between the vectors is \( \cos^{-1}\left( \frac{7}{5\sqrt{2}} \right) \).

Step 1: Understand the Vectors

Given two vectors: \( \mathbf{v_1} = \mathbf{i} + \mathbf{j} \) and \( \mathbf{v_2} = 3\mathbf{i} + 4\mathbf{j} \). We can write them as \( \mathbf{v_1} = (1, 1) \) and \( \mathbf{v_2} = (3, 4) \).

Step 2: Find the Dot Product

The dot product of \( \mathbf{v_1} \) and \( \mathbf{v_2} \) is calculated as follows: \[ \mathbf{v_1} \cdot \mathbf{v_2} = (1)(3) + (1)(4) = 3 + 4 = 7 \]

Step 3: Calculate Magnitude of Each Vector

The magnitude of \( \mathbf{v_1} \) is: \[ ||\mathbf{v_1}|| = \sqrt{1^2 + 1^2} = \sqrt{2} \] The magnitude of \( \mathbf{v_2} \) is: \[ ||\mathbf{v_2}|| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Step 4: Use the Dot Product Formula for Angle

The formula for the angle \( \theta \) between two vectors is: \[ \cos \theta = \frac{\mathbf{v_1} \cdot \mathbf{v_2}}{||\mathbf{v_1}|| \times ||\mathbf{v_2}||} \] Substitute the values: \[ \cos \theta = \frac{7}{\sqrt{2} \times 5} = \frac{7}{5\sqrt{2}} \]

Step 5: Solve for the Angle

To find \( \theta \), compute the inverse cosine:\[ \theta = \cos^{-1}\left( \frac{7}{5\sqrt{2}} \right) \]

Key Concepts

Dot Product

Vectors are fundamental tools in mathematics and physics. One crucial operation involving vectors is the dot product. The dot product is a way to multiply two vectors, resulting in a scalar. It measures how much two vectors are aligned, and gives a sense of their directional relationship.

• The formula for the dot product of two vectors \( \mathbf{v_1} = (a_1, b_1) \) and \( \mathbf{v_2} = (a_2, b_2) \) is:

\[ \mathbf{v_1 \cdot v_2} = a_1 \cdot a_2 + b_1 \cdot b_2 \]
Using this formula, the dot product for our vectors \( \mathbf{v_1} = (1, 1) \) and \( \mathbf{v_2} = (3, 4) \) is calculated as \( 1 \times 3 + 1 \times 4 = 7 \). This scalar result tells us how 'closely' the two vectors point in the same direction - the larger the result, the more aligned they are.

Magnitude of Vectors

Understanding the magnitude of vectors is essential in geometry and physics, as it represents the length or size of the vector itself. The magnitude is computed using the Pythagorean theorem, which provides a way to calculate the length of a vector in its vector space.

• For any vector \( \mathbf{v} = (a, b) \), the magnitude is given by the formula:

\[ ||\mathbf{v}|| = \sqrt{a^2 + b^2} \]
For \( \mathbf{v_1} = (1, 1) \), the magnitude is \( \sqrt{1^2 + 1^2} = \sqrt{2} \). Similarly, for \( \mathbf{v_2} = (3, 4) \), the magnitude is \( \sqrt{3^2 + 4^2} = 5 \). These magnitudes help in assessing the length of each vector. Larger magnitudes signify vectors that have a greater "reach" in the direction they point.

Angle Between Vectors

Calculating the angle between two vectors is crucial in understanding their orientation in a given space. The angle reveals how 'divergent' or 'convergent' the vectors are with respect to each other and is calculated using the cosine of the angle.

• The angle \( \theta \) between two vectors \( \mathbf{v_1} \) and \( \mathbf{v_2} \) is found with the formula:

\[ \cos \theta = \frac{\mathbf{v_1} \cdot \mathbf{v_2}}{||\mathbf{v_1}|| \times ||\mathbf{v_2}||} \]
In our example, substituting the known values: \( \cos \theta = \frac{7}{\sqrt{2} \times 5} = \frac{7}{5 \sqrt{2}} \). Calculating \( \theta \), we use the inverse cosine: \( \theta = \cos^{-1}\left( \frac{7}{5\sqrt{2}} \right) \). This computation gives the precise angle, confirming how much one needs to "rotate" one vector to align it with the other. The angle provides a quantitative measure of direction.