Problem 18 - Precalculus

Question

Find the angle (round to the nearest degree) between each pair of vectors. $$\langle 1,5\rangle \text { and }\langle-3,-2\rangle$$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 135 degrees.

Step 1: Understand the Formula

The angle \( \theta \) between two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) can be found using the formula \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \), where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product and \( \|\mathbf{a}\| \), \( \|\mathbf{b}\| \) are the magnitudes of the vectors.

Step 2: Calculate the Dot Product

Calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \) for the vectors \( \langle 1, 5 \rangle \) and \( \langle -3, -2 \rangle \). The dot product is given by:\[ \mathbf{a} \cdot \mathbf{b} = (1)(-3) + (5)(-2) = -3 - 10 = -13 \]

Step 3: Calculate the Magnitudes of the Vectors

Find the magnitude of each vector. For \( \mathbf{a} = \langle 1, 5 \rangle \):\[ \|\mathbf{a}\| = \sqrt{1^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26} \]For \( \mathbf{b} = \langle -3, -2 \rangle \):\[ \|\mathbf{b}\| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \]

Step 4: Apply the Cosine Formula

Use the results from the previous steps in the cosine formula:\[ \cos \theta = \frac{-13}{\sqrt{26} \times \sqrt{13}} \]Calculate \( \sqrt{26} \times \sqrt{13} = \sqrt{338} \), so:\[ \cos \theta = \frac{-13}{\sqrt{338}} \approx -0.706 \]

Step 5: Calculate the Angle

Use the inverse cosine function to find the angle \( \theta \):\[ \theta = \cos^{-1}(-0.706) \approx 135^{\circ} \]

Key Concepts

dot product

The dot product is a fundamental operation in vector algebra. It involves two vectors and results in a scalar (numerical value). The formula to calculate the dot product of two 2D vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) is:

• \( \mathbf{a} \cdot \mathbf{b} = a_1 \cdot b_1 + a_2 \cdot b_2 \)

In the provided exercise, for the vectors \( \langle 1, 5 \rangle \) and \( \langle -3, -2 \rangle \), we compute the dot product as follows:
- First, multiply corresponding components: - \( 1 \times (-3) \rightarrow -3 \) - \( 5 \times (-2) \rightarrow -10 \)- Then sum these products: - \(-3 + (-10) = -13\)
This operation is significant because it lets us relate the vectors in terms of how much they "point" in the same direction. When dealing with angles between vectors, the dot product forms the numerator of the cosine formula used to find the angle.

vector magnitude

The magnitude of a vector, also known as its length, is a measure of how "long" the vector is from its origin point to its endpoint in a coordinate system. Calculating vector magnitudes is a key step in finding the angle between vectors. The formula for the magnitude of a 2D vector \( \mathbf{a} = \langle a_1, a_2 \rangle \) is:

• \( \|\mathbf{a}\| = \sqrt{a_1^2 + a_2^2} \)

In the exercise, the magnitude calculations are:- For vector \( \mathbf{a} = \langle 1, 5 \rangle \): - \( \|\mathbf{a}\| = \sqrt{1^2 + 5^2} = \sqrt{26} \)- For vector \( \mathbf{b} = \langle -3, -2 \rangle \): - \( \|\mathbf{b}\| = \sqrt{(-3)^2 + (-2)^2} = \sqrt{13} \)Vector magnitudes allow us to find the "size" of vectors regardless of their direction. They are essential for figuring out the denominator part of the cosine formula needed for angle calculations. Once the magnitudes are found, you can use them in conjunction with the dot product to explore further relationships between vectors.

angle between vectors

To find the angle between two vectors, we use a formula that involves the dot product and magnitudes of both vectors. This is commonly done using the relation:

• \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \)

This formula reveals that the cosine of the angle (\( \theta \)) can be determined using the two vectors' dot product and magnitudes.
In our case:- The dot product \( \mathbf{a} \cdot \mathbf{b} \) is \(-13\) as calculated previously.
- Magnitudes are \( \|\mathbf{a}\| = \sqrt{26} \) and \( \|\mathbf{b}\| = \sqrt{13} \).
Insert these into the formula:
- \( \cos \theta = \frac{-13}{\sqrt{26} \times \sqrt{13}} \)- This simplifies to \( \cos \theta \approx -0.706 \)
Finding the angle between vectors is helpful in numerous applications, such as physics and engineering, where understanding vector orientation relative to each other is crucial.

inverse cosine function

Once the cosine of an angle has been identified through calculations such as those earlier described, applying the inverse cosine function (denoted as \( \cos^{-1} \)) helps find the actual angle. The inverse cosine function is the operation that returns the angle \( \theta \) whose cosine is a specified value.In this exercise:- We found \( \cos \theta \approx -0.706 \).
- To get the angle \( \theta \), compute \( \theta = \cos^{-1}(-0.706) \).- This yields \( \theta \approx 135^{\circ} \).Inverse trigonometric functions like inverse cosine are crucial tools since they convert trigonometric ratio outputs back into angle measures. It allows translating geometric and algebraic information into meaningful directional elements, enhancing comprehension of vector relationships.