Problem 45 - University Physics with Modern Physics

Question

Find the angle between each of these pairs of vectors: (a) \(\overrightarrow{A}\) = \(-\)2.00\(\hat{\imath}\) \(+\) 6.00\(\hat{\jmath}\) and \(\overrightarrow{B}\) = 2.00\(\hat{\imath}\) \(-\) 3.00\(\hat{\jmath}\) (b) \(\overrightarrow{A}\) = 3.00\(\hat{\imath}\) \(+\) 5.00\(\hat{\jmath}\) and \(\overrightarrow{B}\) = 10.00\(\hat{\imath}\) \(+\) 6.00\(\hat{\jmath}\) (c) \(\overrightarrow{A}\) = \(-\)4.00\(\hat{\imath}\) \(+\) 2.00\(\hat{\jmath}\) and \(\overrightarrow{B}\) = 7.00\(\hat{\imath}\) \(+\) 14.00\(\hat{\jmath}\)

Step-by-Step Solution

Answer

The angles are approximately 143.1° for (a), 40.4° for (b), and 90° for (c).

Step 1: Use Dot Product Formula

To find the angle \(\theta\) between the vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\), use the dot product formula: \( \overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos(\theta) \). Rearranging this gives \( \cos(\theta) = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|} \).

Step 2(a): Calculate Dot Product for Part (a)

For \(\overrightarrow{A} = -2.00\hat{\imath} + 6.00\hat{\jmath}\) and \(\overrightarrow{B} = 2.00\hat{\imath} - 3.00\hat{\jmath}\), calculate the dot product: \( (-2.00)(2.00) + (6.00)(-3.00) = -4.00 - 18.00 = -22.00 \).

Step 3(a): Calculate Magnitudes for Part (a)

Calculate the magnitudes: \( |\overrightarrow{A}| = \sqrt{(-2.00)^2 + (6.00)^2} = \sqrt{4 + 36} = \sqrt{40} \)\( |\overrightarrow{B}| = \sqrt{(2.00)^2 + (-3.00)^2} = \sqrt{4 + 9} = \sqrt{13} \)

Step 4(a): Find Angle for Part (a)

Substitute into the formula: \( \cos(\theta) = \frac{-22.00}{\sqrt{40} \cdot \sqrt{13}} \)\( \theta = \cos^{-1}\left(\frac{-22.00}{\sqrt{40} \cdot \sqrt{13}}\right) \)Calculate the inverse cosine to find \(\theta\).

Step 2(b): Calculate Dot Product for Part (b)

For \(\overrightarrow{A} = 3.00\hat{\imath} + 5.00\hat{\jmath}\) and \(\overrightarrow{B} = 10.00\hat{\imath} + 6.00\hat{\jmath}\), calculate the dot product: \( (3.00)(10.00) + (5.00)(6.00) = 30.00 + 30.00 = 60.00 \).

Step 3(b): Calculate Magnitudes for Part (b)

Calculate the magnitudes: \( |\overrightarrow{A}| = \sqrt{(3.00)^2 + (5.00)^2} = \sqrt{9 + 25} = \sqrt{34} \)\( |\overrightarrow{B}| = \sqrt{(10.00)^2 + (6.00)^2} = \sqrt{100 + 36} = \sqrt{136} \)

Step 4(b): Find Angle for Part (b)

Substitute into the formula: \( \cos(\theta) = \frac{60.00}{\sqrt{34} \cdot \sqrt{136}} \)\( \theta = \cos^{-1}\left(\frac{60.00}{\sqrt{34} \cdot \sqrt{136}}\right) \)Calculate the inverse cosine to find \(\theta\).

Step 2(c): Calculate Dot Product for Part (c)

For \(\overrightarrow{A} = -4.00\hat{\imath} + 2.00\hat{\jmath}\) and \(\overrightarrow{B} = 7.00\hat{\imath} + 14.00\hat{\jmath}\), calculate the dot product: \( (-4.00)(7.00) + (2.00)(14.00) = -28.00 + 28.00 = 0 \).

Step 3(c): Calculate Magnitudes for Part (c)

Calculate the magnitudes: \( |\overrightarrow{A}| = \sqrt{(-4.00)^2 + (2.00)^2} = \sqrt{16 + 4} = \sqrt{20} \)\( |\overrightarrow{B}| = \sqrt{(7.00)^2 + (14.00)^2} = \sqrt{49 + 196} = \sqrt{245} \)

Step 4(c): Find Angle for Part (c)

Since the dot product is 0, the angle \(\theta\) is 90 degrees (vectors are perpendicular).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is an operation that takes two equal-length sequences of numbers and returns a single number. This is particularly useful in vector mathematics. Given two vectors \(\overrightarrow{A} = a_1\hat{\imath} + a_2\hat{\jmath}\) and \(\overrightarrow{B} = b_1\hat{\imath} + b_2\hat{\jmath}\), the dot product is calculated as follows:

• Multiply the corresponding components: \(a_1 \cdot b_1 + a_2 \cdot b_2\).
• Sum these products to obtain a scalar quantity.

The dot product is very useful in understanding the relationship between two vectors because it combines their magnitudes and the cosine of the angle between them. If the dot product is positive, the vectors point in similar directions, while a negative dot product indicates opposition. A dot product of zero suggests that the vectors are perpendicular.

Vector Magnitudes

The magnitude of a vector is essentially its length or size and is expressed in units of distance. Calculating it involves using the Pythagorean theorem.For a vector \(\overrightarrow{A} = a_1\hat{\imath} + a_2\hat{\jmath}\), its magnitude \(|\overrightarrow{A}|\) is calculated as:

• \(\sqrt{a_1^2 + a_2^2}\), where \(a_1\) and \(a_2\) are the vector components.

Finding the magnitude is crucial because it allows us to understand the scale or extent of the vector irrespective of its direction. Furthermore, when paired with another vector's magnitude in calculations, it helps determine the included angle and assess interactions.

Angle Between Vectors

Calculating the angle between two vectors involves using the dot product and magnitudes of these vectors. The formula for the angle \(\theta\) between two vectors \(\overrightarrow{A}\) and \(\overrightarrow{B}\) is given by:

• \(\cos(\theta) = \frac{\overrightarrow{A} \cdot \overrightarrow{B}}{|\overrightarrow{A}| |\overrightarrow{B}|}\)

This equation shows how to relate the angle to the dot product and magnitudes.Some special cases in angle calculations:

• If the dot product is zero, the angle is \(90^\circ\), indicating perpendicular vectors.
• A \(\cos(\theta)\) value of 1 means the vectors are parallel.
• If \(\cos(\theta)\) is \(-1\), vectors are in opposite directions.

The angle provides insights into how vectors interact spatially.

Cosine Inverse Calculation

Once you have calculated \(\cos(\theta)\), the next step is determining \(\theta\) itself. This is done using the inverse cosine function, often represented as \(\cos^{-1}\).To compute \(\theta\), you must:

• Substitute the calculated \(\cos(\theta)\) value into the inverse cosine function: \(\theta = \cos^{-1}(\cos(\theta))\).

The inverse cosine returns the angle in radians or degrees, depending on your calculator settings.It is essential to note that the inverse cosine function only gives values from \(0^\circ\) to \(180^\circ\), reflecting the possible range of angles between vectors in 2D. Learning to use this function effectively helps you in various vector-related problems, adding a powerful tool to your mathematical toolkit.