Problem 36 - Algebra and Trigonometry

Question

Two vectors u and v are given. Find the angle (expressed in degrees) between u and v. $$ \mathbf{u}=\mathbf{i}+2 \mathbf{j}-2 \mathbf{k}, \quad \mathbf{v}=4 \mathbf{i}-3 \mathbf{k} $$

Step-by-Step Solution

Answer

The angle between the vectors is approximately 48.19 degrees.

Step 1: Write Down the Formula

To find the angle between two vectors \( \mathbf{u} \) and \( \mathbf{v} \), use the formula: \( \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \), where \( \theta \) is the angle, \( \cdot \) represents the dot product, and \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \) are the magnitudes of the vectors.

Step 2: Calculate the Dot Product

Calculate the dot product of \( \mathbf{u} \) and \( \mathbf{v} \). The dot product is \( \mathbf{u} \cdot \mathbf{v} = (1)(4) + (2)(0) + (-2)(-3) = 4 + 0 + 6 = 10 \).

Step 3: Find the Magnitude of Each Vector

Find the magnitudes \( \|\mathbf{u}\| \) and \( \|\mathbf{v}\| \). For \( \mathbf{u} \), \( \| \mathbf{u} \| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \). For \( \mathbf{v} \), \( \| \mathbf{v} \| = \sqrt{4^2 + 0^2 + (-3)^2} = \sqrt{16 + 0 + 9} = \sqrt{25} = 5 \).

Step 4: Plug Values into the Formula

Substitute the dot product and magnitudes into the formula: \( \cos \theta = \frac{10}{3 \times 5} = \frac{10}{15} = \frac{2}{3} \).

Step 5: Calculate the Angle

Use the inverse cosine function to find \( \theta \). \( \theta = \cos^{-1}\left(\frac{2}{3}\right) \). Calculate \( \theta \) using a calculator set to degree mode to get \( \theta \approx 48.19^\circ \).

Key Concepts

Dot Product

The dot product is a fundamental operation in vector algebra, allowing us to multiply two vectors to yield a scalar. It serves as a measure of their directional alignment. When vectors are given in component form,

• written as \( \mathbf{u} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \)
• and \( \mathbf{v} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \).

The dot product is calculated by multiplying corresponding components of the two vectors and adding these products together:\[ \mathbf{u} \cdot \mathbf{v} = ax + by + cz \]The dot product is zero when the vectors are orthogonal (perpendicular), indicating no directional alignment. For the vectors \( \mathbf{u} = \mathbf{i} + 2 \mathbf{j} - 2 \mathbf{k} \) and \( \mathbf{v} = 4 \mathbf{i} - 3 \mathbf{k} \), the calculation was:\[ \mathbf{u} \cdot \mathbf{v} = (1 \times 4) + (2 \times 0) + ((-2) \times (-3)) = 4 + 0 + 6 = 10 \]This implies that the vectors partially align, as denoted by their nonzero dot product.
Understanding the dot product is essential in calculating angles between vectors.

Vector Magnitude

Vector magnitude is akin to the length of a vector in space, denoting how far it extends. Calculating the magnitude involves using the components of the vector and applying the Euclidean distance formula:\[ \| \mathbf{u} \| = \sqrt{a^2 + b^2 + c^2} \]For a vector \( \mathbf{u} = a \mathbf{i} + b \mathbf{j} + c \mathbf{k} \), the magnitude provides a single positive scalar value.
In the example,

• for \( \mathbf{u} = \mathbf{i} + 2 \mathbf{j} - 2 \mathbf{k} \),
• the magnitude is \( \| \mathbf{u} \| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3 \).
• For \( \mathbf{v} = 4 \mathbf{i} - 3 \mathbf{k} \),
• it is \( \| \mathbf{v} \| = \sqrt{4^2 + 0^2 + (-3)^2} = \sqrt{25} = 5 \).

These magnitudes provide necessary scalar measures to assess the directional relationship between vectors using the dot product. A larger magnitude suggests a vector reaching further in space.
It is especially crucial in scenarios, such as physics or engineering, where vector lengths distinguish the scope of forces or displacements.

Inverse Cosine

The inverse cosine function, commonly seen as \( \cos^{-1} \) or \( \text{arccos} \), plays a vital role in determining angles from cosine values. When angles between vectors are required, this trigonometric tool becomes indispensable. By applying the inverse cosine to a known cosine value, you can extract an angle in degrees or radians.
In our vector angle calculation,

• we first find the cosine of the angle using the formula:

\[ \cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|} \]With the dot product and magnitudes, we calculated \( \cos \theta = \frac{2}{3} \).
Applying the inverse cosine,

• \( \theta = \cos^{-1} \left(\frac{2}{3}\right) \)
• gave rise to the angle \( \theta \approx 48.19^\circ \).

This entire process showcases the power of using inverse trigonometric functions in any scenario involving vector angles, widely applicable across scientific and mathematical fields.