Problem 52 - Trigonometry

Question

Find the angle between \(\mathbf{U}=3 \mathbf{i}+7 \mathbf{j}\) and \(\mathbf{V}=\mathbf{i}-4 \mathbf{j}\). a. \(52.8^{\circ}\) b. \(142.8^{\circ}\) c. \(137.2^{\circ}\) d. \(157.9^{\circ}\)

Step-by-Step Solution

Answer

The angle between the vectors is approximately \(142.8^{\circ}\).

Step 1: Calculate Dot Product

The angle between two vectors can be calculated using the dot product formula. The dot product between \(\mathbf{U}\) and \(\mathbf{V}\) is calculated as: \[ \mathbf{U} \cdot \mathbf{V} = 3 \times 1 + 7 \times (-4) = 3 - 28 = -25 \]

Step 2: Calculate Magnitudes

Calculate the magnitudes of vectors \(\mathbf{U}\) and \(\mathbf{V}\): - Magnitude of \(\mathbf{U}\) is: \[ \|\mathbf{U}\| = \sqrt{3^2 + 7^2} = \sqrt{9 + 49} = \sqrt{58} \] - Magnitude of \(\mathbf{V}\) is: \[ \|\mathbf{V}\| = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \]

Step 3: Plug into Cosine Formula

The cosine of the angle \(\theta\) between the two vectors is given by:\[ \cos \theta = \frac{\mathbf{U} \cdot \mathbf{V}}{\|\mathbf{U}\| \cdot \|\mathbf{V}\|} = \frac{-25}{\sqrt{58} \cdot \sqrt{17}} \]

Step 4: Calculate the Cosine Value

Calculate the value of\( \cos \theta \): \[ \cos \theta = \frac{-25}{\sqrt{986}} \approx \frac{-25}{31.4013} \approx -0.796 \]

Step 5: Find the Angle

Find \(\theta\) by taking the inverse cosine of \(-0.796\):\[ \theta \approx \cos^{-1}(-0.796) \approx 142.8^{\circ} \]

Key Concepts

Dot Product

The dot product is a fundamental operation for vectors, which can tell us a great deal about their relationship. It involves multiplying the corresponding components of two vectors and then adding these products together. For example, if we have vector \( \mathbf{U} = a \mathbf{i} + b \mathbf{j} \) and vector \( \mathbf{V} = c \mathbf{i} + d \mathbf{j} \), the dot product \( \mathbf{U} \cdot \mathbf{V} \) is calculated as:

• \( a \times c \)
• + \( b \times d \)

This operation is not only useful for finding the angle between vectors, but also for determining if they are perpendicular. If their dot product is zero, this indicates the vectors are orthogonal. Understanding the dot product is essential for diving deeper into vector mathematics. It simplifies many real-world applications where relationships between directions and forces are key.

Magnitude of Vectors

The magnitude, or length, of a vector is a measure of how "long" it is. It provides a quantifiable scale to the vector, helping us understand its size relative to others. For a vector \( \mathbf{U} = a \mathbf{i} + b \mathbf{j} \), the magnitude \( \|\mathbf{U}\| \) is calculated using the Pythagorean theorem:

• \( \|\mathbf{U}\| = \sqrt{a^2 + b^2} \)

It is similar to finding the hypotenuse of a right triangle with sides \( a \) and \( b \). The magnitude is always a non-negative value and provides crucial information when comparing vectors. In problems involving angles between vectors, knowing the magnitude is key in the cosine formula, connecting size and orientation.

Inverse Cosine

The inverse cosine, denoted as \( \cos^{-1} \), is a trigonometric function used to determine the angle whose cosine is a given number. It is particularly useful when wanting to find the angle between vectors once we have calculated \( \cos \theta \). In practical use, once we have \( \cos \theta \) from our calculations, the inverse cosine allows us to "reverse" the cosine process to get back to the angle \( \theta \).

• This is how we calculate angles from cosine values.
• For angles situated between \(-1\) and \(1\), \( \cos^{-1} \) provides an answer in degrees or radians.

In vector problems like our exercise, \( \cos^{-1} \) lets us translate the cosine relationship into a tangible angle, aiding us in understanding the vectors' directional relationship.