Problem 28 - Trigonometry

Question

Find the angle \(\theta\) between the given vectors to the nearest tenth of a degree. \(\mathrm{U}=4 \mathrm{i}+5 \mathrm{j}, \mathrm{V}=7 \mathrm{i}-4 \mathrm{j}\)

Step-by-Step Solution

Answer

The angle \(\theta\) between the vectors is approximately \(81.0^\circ\).

Step 1: Dot Product of Vectors

To find the angle between two vectors, we first need to calculate their dot product. The dot product of vectors \( \mathbf{U} = 4\mathbf{i} + 5\mathbf{j} \) and \( \mathbf{V} = 7\mathbf{i} - 4\mathbf{j} \) is given by the formula: \( \mathbf{U} \cdot \mathbf{V} = (4)(7) + (5)(-4) \).Calculating this, we get: \[ 28 - 20 = 8 \]. Therefore, the dot product \( \mathbf{U} \cdot \mathbf{V} = 8 \).

Step 2: Magnitude of Vectors

Next, calculate the magnitude of each vector. The magnitude of a vector \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \) is given by \( \| \mathbf{U} \| = \sqrt{a^2 + b^2} \).For \( \mathbf{U} = 4\mathbf{i} + 5\mathbf{j} \), \( \| \mathbf{U} \| = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \).For \( \mathbf{V} = 7\mathbf{i} - 4\mathbf{j} \), \( \| \mathbf{V} \| = \sqrt{7^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65} \).

Step 3: Use the Angle Formula

The angle \( \theta \) between the vectors can be found using the formula:\[\cos \theta = \frac{\mathbf{U} \cdot \mathbf{V}}{\| \mathbf{U} \| \| \mathbf{V} \|}\]Plug in the values from the previous steps:\[\cos \theta = \frac{8}{\sqrt{41} \times \sqrt{65}}\]

Step 4: Calculate \(\cos \theta\) and Find \(\theta\)

Using the values calculated above, solve for \( \cos \theta \):\[\cos \theta = \frac{8}{\sqrt{41} \times \sqrt{65}} \approx 0.1562\]Now, use the inverse cosine function to find \( \theta \):\[\theta = \cos^{-1}(0.1562) \approx 81.0^\circ\]

Step 5: Rounding to the Nearest Tenth

Since the requirements request the angle to the nearest tenth of a degree, we already have our result rounded appropriately: \( 81.0^\circ \). The angle \( \theta \) between the vectors is approximately \( 81.0^\circ \).

Key Concepts

Dot Product of Vectors

The dot product is a fundamental concept when dealing with vectors. It is a way of multiplying two vectors to yield a scalar (a single number). When you have two vectors, such as \( \mathbf{U} = 4\mathbf{i} + 5\mathbf{j} \) and \( \mathbf{V} = 7\mathbf{i} - 4\mathbf{j} \), the dot product is calculated by multiplying their respective components and summing the results.
Here's how it works for our given vectors:

• Take the x-components: 4 (from \(\mathbf{U}\)) and 7 (from \(\mathbf{V}\)), so \(4 \times 7 = 28\).
• Take the y-components: 5 (from \(\mathbf{U}\)) and -4 (from \(\mathbf{V}\)), so \(5 \times -4 = -20\).

Add these products together to find the dot product: 28 + (-20) = 8.
In simple terms, the dot product helps us determine how much one vector extends in the direction of another. It's also crucial when calculating the angle between the vectors.

Magnitude of Vectors

The magnitude of a vector is akin to the length of the vector. It tells us how "long" the vector is from the origin to its endpoint.
Consider a vector with components \( a \) and \( b \), such as \( \mathbf{U} = 4\mathbf{i} + 5\mathbf{j} \). The magnitude \( \| \mathbf{U} \| \) is found using the Pythagorean theorem, which gives us:
\[ \| \mathbf{U} \| = \sqrt{a^2 + b^2} \].

• For \( \mathbf{U} \), this becomes \( \sqrt{4^2 + 5^2} = \sqrt{41} \).
• Similarly, for \( \mathbf{V} = 7\mathbf{i} - 4\mathbf{j} \), the magnitude is \( \sqrt{7^2 + (-4)^2} = \sqrt{65} \).

A vector's magnitude is instrumental in finding the angle between two vectors, as it normalizes them. Without knowing these magnitudes, computing exact angles would be challenging.

Inverse Cosine Function

The inverse cosine function, denoted as \( \cos^{-1} \), is used to find the angle whose cosine value is known. This function is essential in vector mathematics when you wish to determine the angle between two vectors.
To find the angle \(\theta\) between two vectors \(\mathbf{U}\) and \(\mathbf{V}\), we use the formula:
\[ \cos \theta = \frac{\mathbf{U} \cdot \mathbf{V}}{\| \mathbf{U} \| \| \mathbf{V} \|} \].
From our problem, we calculated:

• The dot product \(\mathbf{U} \cdot \mathbf{V} = 8\).
• The magnitudes are \(\sqrt{41}\) and \(\sqrt{65}\) respectively.

Thus, \(\cos \theta = \frac{8}{\sqrt{41} \times \sqrt{65}} \approx 0.1562\).
Finally, to find \(\theta\), apply the inverse cosine function:
\[\theta = \cos^{-1}(0.1562)\] resulting in \(\theta \approx 81.0^\circ\) which is the angle we sought.
This spherical function is vital for converting a cosine value into a tangible angle measurement.