Problem 19 - Precalculus

Question

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

Step-by-Step Solution

Answer

The angle between the vectors is 30 degrees.

Step 1: Recall the Formula for the Angle Between Two Vectors

The angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be found using the dot product formula: \[\cos(\theta) = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}\]where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product and \( \|\mathbf{a}\| \) and \( \|\mathbf{b}\| \) are the magnitudes of the vectors.

Step 2: Calculate the Dot Product

For the vectors \( \mathbf{a} = \langle -2, 2\sqrt{3} \rangle \) and \( \mathbf{b} = \langle -\sqrt{3}, 1 \rangle \), the dot product is calculated as:\[\mathbf{a} \cdot \mathbf{b} = (-2)(-\sqrt{3}) + (2\sqrt{3})(1) = 2\sqrt{3} + 2\sqrt{3} = 4\sqrt{3}\]

Step 3: Calculate the Magnitude of Each Vector

Calculate the magnitude of \( \mathbf{a} \):\[\|\mathbf{a}\| = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4\]Calculate the magnitude of \( \mathbf{b} \):\[\|\mathbf{b}\| = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2\]

Step 4: Substitute into the Cosine Formula

Substitute the values into the cosine formula:\[\cos(\theta) = \frac{4\sqrt{3}}{4 \times 2} = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}\]

Step 5: Find the Angle

The value \( \cos(\theta) = \frac{\sqrt{3}}{2} \) corresponds to an angle \( \theta = 30^\circ \). Therefore, the angle between the vectors, when rounded to the nearest degree, is \( 30^\circ \).

Key Concepts

Angle Between Vectors

Calculating the angle between two vectors is an important concept in vector mathematics. This process helps us understand how two vectors relate in space. The angle tells us the extent of direction change needed to align one vector with another. It is particularly valuable in physics for analysis of forces and in computer graphics for rotational transformations. To find this angle, we use the dot product and magnitudes of the vectors combined into a formula derived from trigonometry. This geometric insight bridges algebra and geometry, offering a practical way to evaluate how two directional 'arrows' interact with one another in a defined space.

Dot Product

The dot product is a fundamental operation for vectors, combining algebraic functions to yield a scalar. It signifies how two vectors 'overlap' or project onto one another. Mathematically, if we have two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is calculated as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

The value obtained indicates the extent to which two vectors are aligned:

• If the dot product is positive, the vectors point in similar directions.
• If zero, they are perpendicular (orthogonal).
• If negative, they point in opposing directions.

In the given problem, the dot product \( \mathbf{a} \cdot \mathbf{b} = 4\sqrt{3} \) signifies a level of directional alignment, crucial for determining the angle between vectors.

Vector Magnitudes

Understanding vector magnitudes is crucial, as they represent a vector's length regardless of direction. It essentially measures the distance from a vector's tail to its tip. For a vector \( \mathbf{a} = \langle a_1, a_2 \rangle \), its magnitude is:

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

This formula follows the Pythagorean theorem, offering a way to quantify the size of vectors:

• Magnitude helps in normalizing vectors, reducing them to unit vectors.
• It's indispensable in physics for computing forces, velocities, and more.

In the exercise, vector magnitudes were \( 4 \) and \( 2 \) respectively, crucial inputs for using the cosine formula to find the angle.

Cosine Formula

The cosine formula elegantly ties together dot product, vector magnitudes, and angles. Derived from the exceptionality of right triangles and trigonometric principles, it is expressed as:

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

This formula unravels the angle \( \theta \) between vectors via cosine calculation, acting as a bridge between straightforward multiplication and angular interpretation.

• It is foundational for directional comparisons and transformations.
• Widely used for calculations involving force components and navigation directions.

For our vectors, using the cosine formula, we found \( \cos(\theta) = \frac{\sqrt{3}}{2} \), which gives us \( \theta = 30^\circ \). The formula illustrates the cohesive use of algebra and geometry to solve real-world problems efficiently.