Problem 33 - Algebra and Trigonometry

Question

Two vectors u and v are given. Find the angle (expressed in degrees) between u and v. $$ \mathbf{u}=\langle 2,-2,-1\rangle, \quad \mathbf{v}=\langle 1,2,2\rangle $$

Step-by-Step Solution

Answer

The angle between vectors \( \mathbf{u} \) and \( \mathbf{v} \) is approximately \( 116.57^{\circ} \).

Step 1: Find the Dot Product

The dot product of two vectors \( \mathbf{u} = \langle a, b, c \rangle \) and \( \mathbf{v} = \langle x, y, z \rangle \) is given by \( \mathbf{u} \cdot \mathbf{v} = ax + by + cz \). For the given vectors, \( \mathbf{u} = \langle 2, -2, -1 \rangle \) and \( \mathbf{v} = \langle 1, 2, 2 \rangle \), the dot product is calculated as \( 2 \cdot 1 + (-2) \cdot 2 + (-1) \cdot 2 = 2 - 4 - 2 = -4 \).

Step 2: Calculate the Magnitudes

The magnitude of a vector \( \mathbf{u} = \langle a, b, c \rangle \) is given by \( ||\mathbf{u}|| = \sqrt{a^2 + b^2 + c^2} \). For \( \mathbf{u} = \langle 2, -2, -1 \rangle \), \( ||\mathbf{u}|| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \). Similarly, for \( \mathbf{v} = \langle 1, 2, 2 \rangle \), \( ||\mathbf{v}|| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).

Step 3: Use the Dot Product Formula for Cosine

The cosine of the angle \( \theta \) between two vectors is given by the formula \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} \). Substituting the values calculated, \( \cos \theta = \frac{-4}{3 \cdot 3} = \frac{-4}{9} \).

Step 4: Find the Angle in Degrees

To find the angle \( \theta \), take the inverse cosine of \( \frac{-4}{9} \), giving \( \theta = \cos^{-1}\left(\frac{-4}{9}\right) \). Calculating this in degrees, \( \theta \approx 116.57^{\circ} \).

Key Concepts

Dot Product

The dot product is a fundamental operation for working with vectors. It helps you understand the interaction between two vectors in relation to their directions. The formal definition for the dot product is:

• For two vectors \( \mathbf{u} = \langle a, b, c \rangle \) and \( \mathbf{v} = \langle x, y, z \rangle \), the dot product is given by \( \mathbf{u} \cdot \mathbf{v} = ax + by + cz \).

The dot product results in a scalar value, which means it does not produce another vector, but rather a single number.This scalar is particularly useful to understand projections and cosines between vectors. If the dot product is zero, the vectors are perpendicular.
In our exercise, we calculate the dot product of vectors \( \mathbf{u} = \langle 2, -2, -1 \rangle \) and \( \mathbf{v} = \langle 1, 2, 2 \rangle \) to get \(-4\). This indicates a certain orientation angle between the vectors.

Magnitude of a Vector

Magnitude is like the length or size of a vector. It gives you a description of how long the vector is if you imagine it as an arrow drawn from the origin of a graph.To find the magnitude of a vector \( \mathbf{u} = \langle a, b, c \rangle \), use the formula:

• \( ||\mathbf{u}|| = \sqrt{a^2 + b^2 + c^2} \)

This formula involves squaring each component of the vector, summing them, and then taking the square root of that sum.
For our vectors, the magnitudes were calculated as:

• For \( \mathbf{u} \), magnitude is \( 3 \) because \( \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3 \).
• For \( \mathbf{v} \), similarly, the magnitude is \( 3 \), using \( \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \).

Understanding magnitudes is crucial for comparing vector sizes and applying them in various calculations.

Cosine of Angle Between Vectors

The cosine of the angle between two vectors is a key concept to determine how the two vectors relate in terms of direction.The formula for finding the cosine of the angle \( \theta \) between vectors \( \mathbf{u} \) and \( \mathbf{v} \) is:

• \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} \)

This translates the interaction of the vectors into a normalized scalar, which can be used to determine the actual angle between them.
In our example, we found \( \cos \theta = \frac{-4}{3 \cdot 3} = \frac{-4}{9} \).This value can then be used to find \( \theta \) using the inverse cosine function, resulting in approximately \( 116.57^{\circ} \).
This angle is critical for analyses in physics and engineering, where understanding how vectors interact in space is essential.