Problem 30 - Trigonometry

Question

Find the angle \(\theta\) between the given vectors to the nearest tenth of a degree. \(\mathbf{U}=11 \mathbf{i}+7 \mathbf{j}, \mathbf{V}=-14 \mathbf{i}+6 \mathbf{j}\)

Step-by-Step Solution

Answer

The angle \( \theta \) is approximately \(124.4^\circ\).

Step 1: Find the Dot Product

To find the angle between two vectors, we first compute their dot product. The dot product of two vectors \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \) and \( \mathbf{V} = c\mathbf{i} + d\mathbf{j} \) is given by: \[ \mathbf{U} \cdot \mathbf{V} = ac + bd \] For the vectors \( \mathbf{U} = 11\mathbf{i} + 7\mathbf{j} \) and \( \mathbf{V} = -14\mathbf{i} + 6\mathbf{j} \), the dot product is: \[ 11 \times (-14) + 7 \times 6 = -154 + 42 = -112 \]

Step 2: Calculate the Magnitudes of the Vectors

Next, we need the magnitudes of \( \mathbf{U} \) and \( \mathbf{V} \). The magnitude of a vector \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \) is:\[ ||\mathbf{U}|| = \sqrt{a^2 + b^2} \]For \( \mathbf{U} \):\[ ||\mathbf{U}|| = \sqrt{11^2 + 7^2} = \sqrt{121 + 49} = \sqrt{170} \]And for \( \mathbf{V} \):\[ ||\mathbf{V}|| = \sqrt{(-14)^2 + 6^2} = \sqrt{196 + 36} = \sqrt{232} \]

Step 3: Apply the Cosine Formula to Find the Angle

The cosine of the angle \( \theta \) between two vectors is given by the formula:\[ \cos \theta = \frac{\mathbf{U} \cdot \mathbf{V}}{||\mathbf{U}|| \times ||\mathbf{V}||} \]Substitute the values we found:\[ \cos \theta = \frac{-112}{\sqrt{170} \times \sqrt{232}} = \frac{-112}{\sqrt{170 \times 232}} \]

Step 4: Compute the Cosine Value

Calculate the denominator from Step 3:\[ \sqrt{170 \times 232} \approx \sqrt{39440} \approx 198.6 \] Thus, \[ \cos \theta \approx \frac{-112}{198.6} \approx -0.564 \]

Step 5: Find the Angle Using Inverse Cosine

The angle \( \theta \) can be found by taking the inverse cosine (arccos) of \( -0.564 \):\[ \theta = \cos^{-1}(-0.564) \approx 124.4^\circ \]

Key Concepts

Dot Product

The dot product is a crucial operation when dealing with vectors, especially in finding the angle between them. Imagine the dot product as a way to combine two vectors into a single numerical value that helps us understand how much they point in the same direction. It's like checking how aligned two paths are.

To calculate the dot product of two vectors, you multiply their corresponding components and then sum those products. For example, if you have vectors \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \) and \( \mathbf{V} = c\mathbf{i} + d\mathbf{j} \), then their dot product is given by the expression:

• \( \mathbf{U} \cdot \mathbf{V} = ac + bd \)

In the problem provided, vectors \( \mathbf{U} \) and \( \mathbf{V} \) are:

• \( \mathbf{U} = 11\mathbf{i} + 7\mathbf{j} \)
• \( \mathbf{V} = -14\mathbf{i} + 6\mathbf{j} \)

The dot product computes as:

• \( 11 \times (-14) + 7 \times 6 = -154 + 42 = -112 \)

This value, \(-112\), tells us something vital about the vectors' alignment, which is used next to find the angle between them.

Vector Magnitudes

Understanding the magnitude of a vector is like figuring out how long it is or how much ground it covers. It's a reflection of the vector's "size." Calculating magnitude is one of the basic operations in vector analysis and crucial for determining angles between vectors.

• The magnitude |\( \mathbf{U} \)| of a vector \( \mathbf{U} = a\mathbf{i} + b\mathbf{j} \) is calculated as:

\[||\mathbf{U}|| = \sqrt{a^2 + b^2}\]To put it in simple terms, you're applying the Pythagorean theorem to the vector's components, where \( a \) and \( b \) are like the sides of a right triangle, and \( ||\mathbf{U}|| \) is the hypotenuse.

For the vectors in the exercise:

• \( ||\mathbf{U}|| = \sqrt{11^2 + 7^2} = \sqrt{121 + 49} = \sqrt{170} \)
• \( ||\mathbf{V}|| = \sqrt{(-14)^2 + 6^2} = \sqrt{196 + 36} = \sqrt{232} \)

Knowing the magnitude of each vector is a step toward finding the angle between them, as it helps us normalize the vectors and use them in the cosine formula.

Cosine of Angle Between Vectors

Once you have the dot product and the magnitudes of two vectors, you can find the cosine of the angle between them using a straightforward formula. This is like measuring how "open" the angle between two paths is if you were to stand at the origin.

• The formula to find the cosine of the angle \( \theta \) between vectors \( \mathbf{U} \) and \( \mathbf{V} \) is:

\[\cos \theta = \frac{\mathbf{U} \cdot \mathbf{V}}{||\mathbf{U}|| \times ||\mathbf{V}||}\]This indicates that the cosine relates the dot product of the vectors to their magnitudes.

In the context of the sample problem, having calculated a dot product of \(-112\), and magnitudes \(\sqrt{170}\) and \(\sqrt{232}\), you plug these into the cosine formula:

• \( \cos \theta = \frac{-112}{\sqrt{170} \times \sqrt{232}} \approx \frac{-112}{198.6} \approx -0.564 \)

Finally, to find the angle itself, you take the inverse cosine (arccos) of \(-0.564\), revealing the angle's actual size.
Taking the inverse cosine gives you:

• \( \theta = \cos^{-1}(-0.564) \approx 124.4^\circ \)

Thus, you find that the angle between these vectors is approximately \( 124.4^\circ \), indicating that they are pointing relatively away from each other but not completely opposed.