Problem 8 - Precalculus: Functions and Graphs

Question

Find (a) the dot product of the two vectors and (b) the angle between the two vectors. $$\langle- 3,6\rangle, \quad\langle- 1,2\rangle$$

Step-by-Step Solution

Answer

(a) The dot product is 15. (b) The angle between the vectors is 0°.

Step 1: Identify the two vectors

We have two vectors: \( \vec{A} = \langle -3, 6 \rangle \) and \( \vec{B} = \langle -1, 2 \rangle \).

Step 2: Calculate the dot product

The dot product of two vectors \( \vec{A} = \langle a_1, a_2 \rangle \) and \( \vec{B} = \langle b_1, b_2 \rangle \) is given by the formula: \( \vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 \). Substituting the given values, we calculate: \( (-3)(-1) + (6)(2) = 3 + 12 = 15 \). Thus, the dot product is 15.

Step 3: Calculate the magnitudes of the vectors

The magnitude of a vector \( \vec{A} = \langle a_1, a_2 \rangle \) is given by \( ||\vec{A}|| = \sqrt{a_1^2 + a_2^2} \). For \( \vec{A} \), we have: \( ||\vec{A}|| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} \). Simplifying gives \( \sqrt{45} = 3\sqrt{5} \).

Step 4: Calculate magnitude of the second vector

For \( \vec{B} = \langle -1, 2 \rangle \), the magnitude is calculated as: \( ||\vec{B}|| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \).

Step 5: Use the dot product formula to find the angle

The formula relating the dot product to the angle \( \theta \) between two vectors is: \( \vec{A} \cdot \vec{B} = ||\vec{A}|| ||\vec{B}|| \cos\theta \). Substituting the known values: \( 15 = (3\sqrt{5})(\sqrt{5}) \cos\theta = 15\cos\theta \).

Step 6: Solve for \( \cos\theta \) and find the angle

Rearrange the equation: \( \cos\theta = \frac{15}{15} = 1 \). Since \( \cos\theta = 1 \), the angle \( \theta = 0 \) degrees. Thus, the vectors are parallel.

Key Concepts

Understanding Vectors

Vectors are fundamental elements in mathematics and physics used to describe quantities having both magnitude and direction. In simpler terms, think of them as arrows pointing from one place to another. These are particularly useful for explaining various physical phenomena, such as velocity or force.
Consider a vector represented as \( \vec{A} = \langle a_1, a_2 \rangle \). Here, \( a_1 \) and \( a_2 \) describe the vector's components along the x-axis and y-axis, respectively. To better visualize, you can imagine it starting from the origin of a graph extending towards a point defined by its components.
Key attributes of vectors include:

• Direction: Shown by the angle the vector makes with a reference axis.
• Magnitude: Its "size" or length.

Magnitude of a Vector

The magnitude of a vector indicates how long or large a vector is. You can think of it as the distance from the origin to the point defined by the vector. Understanding magnitude is key to comparing vectors and calculating other properties like velocity or force.
To find the magnitude of a vector \( \vec{A} = \langle a_1, a_2 \rangle \), it uses the Pythagorean theorem to calculate the length as follows: \[||\vec{A}|| = \sqrt{a_1^2 + a_2^2}\]
For instance, the magnitude of \( \vec{A} = \langle -3, 6 \rangle \) would be:

• \( ||\vec{A}|| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \)

Calculating the magnitude allows us to interpret the "length" component of the vector visually.

Angle Between Vectors

Finding the angle between two vectors involves understanding how they orient with respect to each other in space. This angle is crucial, especially when analyzing resultant vectors or determining how two vectors interact.
The dot product, a scalar, helps in calculating this angle through the following relation:\[\vec{A} \cdot \vec{B} = ||\vec{A}|| ||\vec{B}|| \cos \theta\]Here, \( \theta \) represents the angle between vectors \( \vec{A} \) and \( \vec{B} \). Solving for \( \cos \theta \), you will use the dot product and magnitudes:

• \( \cos \theta = \frac{\vec{A} \cdot \vec{B}}{||\vec{A}|| \cdot ||\vec{B}||} \)

In our example, setting the calculated values gives \( \cos \theta = 1 \), indicating \( \theta = 0 \) degrees. This means the vectors are parallel, pointing in the same direction.

The Role of Trigonometry

Trigonometry bridges the gap between algebra and geometry, providing tools to solve vector-related problems efficiently. By applying trigonometric identities and principles, we can find unknown angles and lengths in vector problems.
In the context of vectors, trigonometry allows us to understand relationships between angles and sides through functions like sine, cosine, and tangent. The cosine formula used with the dot product is a direct application of this:

• \( \cos \theta = \frac{\text{dot product}}{\text{product of magnitudes}} \)

This calculation helps us deduce if vectors are perpendicular, parallel, or forming an angle in between. Whenever vectors are perpendicular, their dot product is zero, representing a 90-degree angle. Parallel vectors, like our exercise demonstrates, result in \( \cos \theta = 1 \).
Thus, trigonometry not only tells how vectors sit relative to each other but also helps model real-world situations requiring direction and magnitude analysis.