Problem 29 - Trigonometry

Question

Find the angle \(\theta\) between the given vectors to the nearest tenth of a degree. \(\mathbf{U}=13 \mathbf{i}-8 \mathbf{j}, \mathbf{V}=2 \mathbf{i}+11 \mathbf{j}\)

Step-by-Step Solution

Answer

The angle between the vectors is approximately \( 111.2^\circ \).

Step 1: Recall the formula for the angle between two vectors

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

Step 2: Compute the dot product of the vectors

The dot product \( \mathbf{U} \cdot \mathbf{V} \) is calculated as follows: \[ \mathbf{U} \cdot \mathbf{V} = (13)(2) + (-8)(11) = 26 - 88 = -62 \]

Step 3: Find the magnitude of each vector

The magnitude of a vector \( \mathbf{U} = a \mathbf{i} + b \mathbf{j} \) is given by \[ \| \mathbf{U} \| = \sqrt{a^2 + b^2} \]. Therefore, the magnitudes are: \( \| \mathbf{U} \| = \sqrt{13^2 + (-8)^2} = \sqrt{169 + 64} = \sqrt{233} \approx 15.26 \) \( \| \mathbf{V} \| = \sqrt{2^2 + 11^2} = \sqrt{4 + 121} = \sqrt{125} = 11.18 \)

Step 4: Substitute the known values into the formula

Now, substitute the values into the cosine formula: \[ \cos \theta = \frac{-62}{15.26 \times 11.18} = \frac{-62}{170.65} \approx -0.363 \]

Step 5: Calculate the angle \( \theta \) using inverse cosine

Find \( \theta \) using the arccosine function: \[ \theta = \cos^{-1}(-0.363) \approx 111.2^\circ \] Thus, the angle between the vectors is approximately \( 111.2^\circ \).

Key Concepts

Dot Product

To solve problems involving the angle between vectors, understanding the dot product is crucial. The dot product is a way to multiply two vectors, giving a scalar as a result. This operation helps measure how much one vector extends in the direction of another vector.
The formula for the dot product of two vectors \(\mathbf{U}\) and \(\mathbf{V}\) is:

• \( \mathbf{U} \cdot \mathbf{V} = a_1b_1 + a_2b_2 \)

where \(\mathbf{U} = a_1\mathbf{i} + a_2\mathbf{j}\) and \(\mathbf{V} = b_1\mathbf{i} + b_2\mathbf{j}\).
For our vectors \(\mathbf{U} = 13\mathbf{i} - 8\mathbf{j}\) and \(\mathbf{V} = 2\mathbf{i} + 11\mathbf{j}\), using the formula we have:

• \( \mathbf{U} \cdot \mathbf{V} = 13 \times 2 + (-8) \times 11 = 26 - 88 = -62 \)

This result indicates that the vectors have an angle larger than 90 degrees between them.

Vector Magnitude

The magnitude of a vector, often referred to as the vector's length, is critical for determining the angle between vectors. The magnitude is a measure of how long the vector is, similar to the length of a line segment in geometry.
It is calculated using the Pythagorean theorem in a two-dimensional coordinate system.

• For vector \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \), the magnitude \( \|\mathbf{U}\| \) is:
• \( \|\mathbf{U}\| = \sqrt{a^2 + b^2} \)

Applying this formula to the vectors \(\mathbf{U} = 13\mathbf{i} - 8\mathbf{j}\) and \(\mathbf{V} = 2\mathbf{i} + 11\mathbf{j}\):

• \( \|\mathbf{U}\| = \sqrt{13^2 + (-8)^2} = \sqrt{169 + 64} = \sqrt{233} \approx 15.26 \)
• \( \|\mathbf{V}\| = \sqrt{2^2 + 11^2} = \sqrt{4 + 121} = \sqrt{125} = 11.18 \)

These magnitudes are required to determine how the vectors relate to each other in space.

Inverse Cosine

Once we have calculated the dot product and the magnitudes of the vectors, the next step in finding the angle between them involves using the inverse cosine function, often denoted as \( \cos^{-1} \) or arccosine.
This function helps us retrieve an angle from a cosine value.

• The formula for the angle \( \theta \) between two vectors \(\mathbf{U}\) and \(\mathbf{V}\) is:
• \( \theta = \cos^{-1}\left( \frac{ \mathbf{U} \cdot \mathbf{V} }{ \| \mathbf{U} \| \| \mathbf{V} \| } \right) \)

Inserting our computed values:

• \( \cos \theta = \frac{-62}{15.26 \times 11.18} \approx -0.363 \)
• \( \theta = \cos^{-1}(-0.363) \approx 111.2^\circ \)

This result shows the angle between vectors is about \( 111.2^\circ \). Each value and step are essential in assessing the direction and orientation relationship between the vectors.

Trigonometry Problem

Connecting all these concepts forms a complete trigonometry problem where you are asked to determine the angle by utilizing vector operations. Understanding how vector mathematics relates to trigonometry helps solve problems in physics and engineering that involve forces and directions.
This approach applies the fundamental trigonometric identity involving cosine to vector analysis, combining linear algebra with trigonometric functions.

• Steps we followed include:
• Calculating the dot product to measure direction alignment.
• Finding the magnitudes to understand vector lengths.
• Using inverse cosine for determining the angle.

Such problems show how trigonometry is not just confined to triangles, but expands into understanding complex spatial relationships between forces, velocities, and more in a coordinate system.