Problem 55 - University Physics with Modern Physics

Question

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

Step-by-Step Solution

Answer

(a) 162.25°, (b) 27.17°, and (c) 90°.

Step 1: Calculate Dot Product (a)

Given vectors \(\vec{A} = -2.00 \hat{\imath} + 6.00 \hat{\jmath}\) and \(\vec{B} = 2.00 \hat{\imath} - 3.00 \hat{\jmath}\), the dot product is calculated as: \(\vec{A} \cdot \vec{B} = (-2.00)(2.00) + (6.00)(-3.00) = -4.00 - 18.00 = -22.00\).

Step 2: Calculate Magnitudes (a)

The magnitude of \(\vec{A}\) is \(|\vec{A}| = \sqrt{(-2.00)^2 + (6.00)^2} = \sqrt{4.00 + 36.00} = \sqrt{40.00} = 6.32\). Likewise, \(|\vec{B}| = \sqrt{(2.00)^2 + (-3.00)^2} = \sqrt{4.00 + 9.00} = \sqrt{13.00} = 3.61\).

Step 3: Calculate Angle (a)

To find the angle \(\theta\) between \(\vec{A}\) and \(\vec{B}\), use the formula: \(\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}\). Thus, \(\cos \theta = \frac{-22.00}{6.32 \times 3.61}\) which gives \(\theta = \cos^{-1}(-0.944) = 162.25^\circ\).

Step 4: Calculate Dot Product (b)

For vectors \(\vec{A} = 3.00 \hat{\imath} + 5.00 \hat{\jmath}\) and \(\vec{B} = 10.00 \hat{\imath} + 6.00 \hat{\jmath}\), the dot product is \(\vec{A} \cdot \vec{B} = (3.00)(10.00) + (5.00)(6.00) = 30.00 + 30.00 = 60.00\).

Step 5: Calculate Magnitudes (b)

The magnitude of \(\vec{A}\) is \(|\vec{A}| = \sqrt{(3.00)^2 + (5.00)^2} = \sqrt{9.00 + 25.00} = \sqrt{34.00} = 5.83\). The magnitude of \(\vec{B}\) is \(|\vec{B}| = \sqrt{(10.00)^2 + (6.00)^2} = \sqrt{100.00 + 36.00} = \sqrt{136.00} = 11.66\).

Step 6: Calculate Angle (b)

Using \(\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|}\), we find \(\cos \theta = \frac{60.00}{5.83 \times 11.66}\) resulting in \(\theta = \cos^{-1}(0.883) = 27.17^\circ\).

Step 7: Calculate Dot Product (c)

For vectors \(\vec{A} = -4.00 \hat{\imath} + 2.00 \hat{\jmath}\) and \(\vec{B} = 7.00 \hat{\imath} + 14.00 \hat{\jmath}\), the dot product is \(\vec{A} \cdot \vec{B} = (-4.00)(7.00) + (2.00)(14.00) = -28.00 + 28.00 = 0\).

Step 8: Calculate Magnitudes (c)

The magnitude of \(\vec{A}\) is \(|\vec{A}| = \sqrt{(-4.00)^2 + (2.00)^2} = \sqrt{16.00 + 4.00} = \sqrt{20.00} = 4.47\). The magnitude of \(\vec{B}\) is \(|\vec{B}| = \sqrt{(7.00)^2 + (14.00)^2} = \sqrt{49.00 + 196.00} = \sqrt{245.00} = 15.65\).

Step 9: Calculate Angle (c)

Since \(\vec{A} \cdot \vec{B} = 0\), vectors \(\vec{A}\) and \(\vec{B}\) are perpendicular, meaning the angle \(\theta = 90^\circ\).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is an essential operation used in vector mathematics. It is used to measure the extent to which two vectors align with each other.
It is calculated by multiplying the corresponding components of the vectors and then summing those products. For example, if we have vectors \(\vec{A} = a_1\hat{\imath} + a_2\hat{\jmath}\) and \(\vec{B} = b_1\hat{\imath} + b_2\hat{\jmath}\), the dot product is computed as \(\vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2\).
This operation results in a single number, which can be positive, negative, or zero.

• A positive dot product indicates that the vectors are pointing in generally the same direction.
• A negative dot product suggests that the vectors are pointing in opposite directions.
• A dot product of zero signifies that the vectors are perpendicular to each other.

Understanding dot products is crucial, especially when calculating angles between vectors, as it reveals directional relationships.

Magnitude of a Vector

The magnitude of a vector is essentially its length or size. It provides a measure of how long the vector is regardless of its direction.
To compute this, you use the Pythagorean theorem. For a vector \(\vec{A} = a_1\hat{\imath} + a_2\hat{\jmath}\), the magnitude is calculated as \(|\vec{A}| = \sqrt{a_1^2 + a_2^2}\). This formula derives from the geometry of right triangles, treating the vector as the hypotenuse.
Let's consider why this is important:

• Magnitude is crucial for scaling vectors, as it allows comparison of different vector lengths.
• It is an important component in calculating unit vectors, which are vectors with a magnitude of one.
• Knowing magnitudes is necessary to find angles between vectors.

In vector calculations, paying attention to magnitudes ensures that we accurately understand and utilize vector properties.

Cosine of Angle Between Vectors

To find the angle between two vectors, we use the concept of cosine, which relates angles with dot products and magnitudes. The formula to calculate the angle \(\theta\) between vectors \(\vec{A}\) and \(\vec{B}\) is \(\cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| \cdot |\vec{B}|}\). This equation uses:

• The dot product which signifies vector alignment.
• The magnitudes to normalize the vectors.

By computing the cosine of the angle, we determine whether the vectors are:

• Parallel (cosine near 1).
• Perpendicular (cosine exactly 0).
• Anti-parallel (cosine near -1).

Once the cosine value is obtained, you find \(\theta\) by calculating the inverse cosine, also known as arccosine.
This gives the angle in degrees or radians, providing a clear picture of the spatial relationship between the vectors. Understanding these calculations is key when analyzing vector alignments in physics and engineering.