Problem 35 - Precalculus with Limits

Question

Finding the Angle Between Two Vectors In Exercises \(31-40,\) find the angle \(\theta\) between the vectors. $$\mathbf{u}=2 \mathbf{i}-\mathbf{j}$$ $$\mathbf{v}=6 \mathbf{i}+4 \mathbf{j}$$

Step-by-Step Solution

Answer

The angle between vectors \(\mathbf{u}\) and \(\mathbf{v}\) is approximately \(60.93°\).

Step 1: Calculate the Dot Product of \(\mathbf{u}\) and \(\mathbf{v}\)

The dot product of two vectors is calculated as the sum of the product of their corresponding components. Therefore, \(\mathbf{u} \cdot \mathbf{v} = (2)(6) + (-1)(4) = 8.

Step 2: Calculate the Magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\)

The magnitude of a vector is found from the square root of the sum of squares of its components. \nTherefore, \(\|\mathbf{u}\| = \sqrt{(2)^2 + (-1)^2} = \sqrt{5}\) and \(\|\mathbf{v}\| = \sqrt{(6)^2 + (4)^2} = \sqrt{52}\).

Step 3: Calculate the angle \(\theta\) using dot product and magnitudes

The angle \(\theta\) between two vectors is given by the formula \(\cos(\theta) = \frac{(\mathbf{u} \cdot \mathbf{v})}{(\|\mathbf{u}\| \|\mathbf{v}\|)}\). Plugging in previously calculated values, \(\cos(\theta) = \frac{8}{\sqrt{5} \sqrt{52}}\), solving this we get \(\cos(\theta) = 0.485\). The angle \(\theta\) is then \(\arccos(0.485) \approx 60.93°\).

Key Concepts

Vector Mathematics

Vector mathematics is all about understanding vectors and how they interact with each other in mathematical terms. Think of a vector as an arrow that points in a particular direction with a certain magnitude or length. Vectors are incredibly useful because they can describe things like forces, velocities, and other physical quantities in multiple dimensions.
**Components and Notation**Vectors are often denoted by bold letters like \( \mathbf{u} \) and \( \mathbf{v} \). They have components that define their direction and magnitude:

• For example, the vector \( \mathbf{u} = 2\mathbf{i} - \mathbf{j} \) has components 2 and -1 along the x and y axes, respectively.
• These components tell us how far the vector moves along each axis.

Combining vectors using operations such as addition, subtraction, and scalar multiplication allows us to find resultant vectors and many other useful calculations.

Dot Product

The dot product is a way to combine two vectors and get a scalar (a plain number), rather than another vector. It's especially useful because it reflects the angle between the two vectors. Here's how it works:**Calculating the Dot Product**For two vectors \( \mathbf{u} \) and \( \mathbf{v} \), each with components, the dot product is given by:\[\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + \ldots\]

• In our example, \( \mathbf{u} \cdot \mathbf{v} = (2)(6) + (-1)(4) = 8 \).
• The result is a single number, 8, which tells us something about how the vectors align.

The dot product is zero if the vectors are perpendicular, greater than zero if they point in the same general direction, and less than zero if they point in opposite directions.

Magnitude of Vectors

The magnitude of a vector is like the length of the arrow in the vector diagram. It's a measure of how strong or intense the vector quantity is.**Calculating the Magnitude**The magnitude of a vector \( \mathbf{u} \) with components \( u_1, u_2, \ldots \) is calculated using the Pythagorean theorem as follows:\[\|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2 + \ldots}\]

• For \( \mathbf{u} = 2\mathbf{i} - \mathbf{j} \), its magnitude is \( \|\mathbf{u}\| = \sqrt{2^2 + (-1)^2} = \sqrt{5} \).
• Similarly, the magnitude of \( \mathbf{v} = 6\mathbf{i} + 4\mathbf{j} \) is \( \|\mathbf{v}\| = \sqrt{6^2 + 4^2} = \sqrt{52} \).

These calculations give us crucial data about the strength or extent of each vector.

Trigonometry in Vectors

Trigonometry allows us to explore the angles and relationships between vectors. In the context of finding the angle between vectors, trigonometry plays a vital role.**Finding the Angle**The cosine of the angle \( \theta \) between two vectors \( \mathbf{u} \) and \( \mathbf{v} \) is calculated using the formula:\[\cos(\theta) = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\]

• In our exercise, \( \cos(\theta) = \frac{8}{\sqrt{5} \sqrt{52}} \approx 0.485 \).
• To find \( \theta \), we use the inverse cosine function, \( \theta = \arccos(0.485) \).
• This gives us an angle of approximately \( 60.93^\circ \).

This mathematical relationship helps visualize how vectors interact, aligning trigonometry with vector operations for practical problem-solving.