Problem 34 - Algebra and Trigonometry

Question

Two vectors u and v are given. Find the angle (expressed in degrees) between u and v. $$ \mathbf{u}=\langle 4,0,2\rangle, \quad \mathbf{v}=\langle 2,-1,0\rangle $$

Step-by-Step Solution

Answer

The angle between vectors \(\mathbf{u}\) and \(\mathbf{v}\) is approximately 36.87 degrees.

Step 1: Understand the Dot Product Formula

To find the angle between two vectors, we need to use the formula \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \ ||\mathbf{v}||} \). First, let's understand what this formula involves. The dot product \( \mathbf{u} \cdot \mathbf{v} \) is calculated as the sum of the products of their corresponding components, and the magnitudes \( ||\mathbf{u}|| \) and \( ||\mathbf{v}|| \) are the lengths of the vectors.

Step 2: Calculate the Dot Product

Find the dot product of \( \mathbf{u} = \langle 4,0,2 \rangle \) and \( \mathbf{v} = \langle 2,-1,0 \rangle \). The dot product is calculated as follows:\( \mathbf{u} \cdot \mathbf{v} = 4 \times 2 + 0 \times (-1) + 2 \times 0 = 8 + 0 + 0 = 8 \).

Step 3: Calculate the Magnitudes of the Vectors

Compute the magnitudes of \( \mathbf{u} \) and \( \mathbf{v} \). The magnitude of \( \mathbf{u} \) is:\( ||\mathbf{u}|| = \sqrt{4^2 + 0^2 + 2^2} = \sqrt{16 + 0 + 4} = \sqrt{20} \). The magnitude of \( \mathbf{v} \) is:\( ||\mathbf{v}|| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1 + 0} = \sqrt{5} \).

Step 4: Apply the Dot Product Formula

Plug the dot product and magnitudes into the cosine formula:\( \cos \theta = \frac{8}{\sqrt{20} \times \sqrt{5}} = \frac{8}{\sqrt{100}} = \frac{8}{10} \). Simplify to get \( \cos \theta = 0.8 \).

Step 5: Find the Angle in Radians

To find \( \theta \), take the inverse cosine of 0.8:\( \theta = \cos^{-1}(0.8) \approx 0.6435 \text{ radians} \).

Step 6: Convert Angle from Radians to Degrees

Convert the angle from radians to degrees using the conversion: \( 1 \text{ radian} = \left( \frac{180}{\pi} \right) \) degrees.\( \theta \approx 0.6435 \times \frac{180}{\pi} \approx 36.87 \text{ degrees} \).

Key Concepts

The Dot Product

The dot product, also known as the scalar product, is a way to multiply two vectors, resulting in a scalar (a number, not a vector). This is distinct from vector multiplication, which results in another vector. Understanding the dot product is pivotal when dealing with problems related to the angle between vectors.
The dot product of vectors \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle\) and \(\mathbf{v} = \langle v_1, v_2, v_3 \rangle\) is calculated as follows:
\[ \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 \]

This multiplication combines corresponding components of each vector, resulting in a single value. For example, with vectors \(\mathbf{u} = \langle 4, 0, 2 \rangle\) and \(\mathbf{v} = \langle 2, -1, 0 \rangle\), the dot product is calculated as:
\[ \mathbf{u} \cdot \mathbf{v} = 4 \times 2 + 0 \times (-1) + 2 \times 0 = 8 + 0 + 0 = 8 \]
The dot product reflects the extent to which two vectors point in the same general direction. A positive dot product indicates a general alignment, while a zero dot product suggests the vectors are perpendicular.

Magnitude of a Vector

The magnitude of a vector, often called its length or norm, is a measure of how long the vector is. Calculating the magnitude is crucial because it allows us to quantify the size of the vector in space.
For a vector \(\mathbf{u} = \langle u_1, u_2, u_3 \rangle\), its magnitude \(||\mathbf{u}||\) is found using the Pythagorean theorem as follows:
\[ ||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2 + u_3^2} \]

This formula effectively extends the Pythagorean theorem to three-dimensional space. Understanding how to compute this helps us further calculate the angle between vectors.
Given \(\mathbf{u} = \langle 4, 0, 2 \rangle\), the magnitude is:
\[ ||\mathbf{u}|| = \sqrt{4^2 + 0^2 + 2^2} = \sqrt{16 + 0 + 4} = \sqrt{20} \]

Similarly, for \(\mathbf{v} = \langle 2, -1, 0 \rangle\):
\[ ||\mathbf{v}|| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{4 + 1 + 0} = \sqrt{5} \]
Magnitudes are always non-negative and give us a universal way to compare vector lengths.

Inverse Cosine for Angle Calculation

Inverse cosine, denoted as \(\cos^{-1}\) or arccos, is a critical trigonometric function that allows us to determine the angle between two vectors given their dot product and magnitudes. It's especially useful when trying to figure out angles in scenarios where only vector information is available.
The formula to find the angle \(\theta\) between two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is:
\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| ||\mathbf{v}||} \]

Once you have calculated this ratio, the inverse cosine function helps convert this value back into an angle. For our given vectors, we've established:
\[ \cos \theta = 0.8 \]
Applying the inverse cosine function provides the angle in radians:
\[ \theta = \cos^{-1}(0.8) \approx 0.6435 \text{ radians} \]
To translate this into degrees, which is often more intuitive, use the conversion:
\[ \theta \approx 0.6435 \times \frac{180}{\pi} \approx 36.87 \text{ degrees} \]
This process ultimately unlocks the angle measurement, facilitating deeper understanding in geometry and physics contexts.