Problem 74 - Fundamentals of Physics

Question

Vector \(\vec{a}\) lies in the \(y z\) plane \(63.0^{\circ}\) from the positive direction of the \(y\) axis, has a positive \(z\) component, and has magnitude \(3.20\) units. Vector \(\vec{b}\) lies in the \(x z\) plane \(48.0^{\circ}\) from the positive direction of the \(x\) axis, has a positive \(z\) component, and has magnitude \(1.40\) units. Find (a) \(\vec{a} \cdot \vec{b},(\) b) \(\vec{a} \times \vec{b}\), and (c) the angle between \(\vec{a}\) and \(\vec{b}\).

Step-by-Step Solution

Answer

(a) 2.964, (b) \( \langle 1.508, -2.679, -1.363 \rangle \), (c) \( 48.69^\circ \)

Step 1: Resolve Vector a into Components

Vector \( \vec{a} \) lies in the \( yz \) plane, so it has no component in the \( x \) direction. Its components are determined by the angle \( 63.0^\circ \) with respect to the \( y \)-axis. Since it has a positive \( z \) component, we have:\[ a_y = 3.20 \cos(63.0^\circ) \approx 1.45 \]\[ a_z = 3.20 \sin(63.0^\circ) \approx 2.85 \]Thus, \( \vec{a} = \langle 0, 1.45, 2.85 \rangle \).

Step 2: Resolve Vector b into Components

Vector \( \vec{b} \) lies in the \( xz \) plane, so it has no component in the \( y \) direction. Its components are determined by the angle \( 48.0^\circ \) with respect to the \( x \)-axis. Since it has a positive \( z \) component, we have:\[ b_x = 1.40 \cos(48.0^\circ) \approx 0.94 \]\[ b_z = 1.40 \sin(48.0^\circ) \approx 1.04 \]Thus, \( \vec{b} = \langle 0.94, 0, 1.04 \rangle \).

Step 3: Calculate the Dot Product \( \vec{a} \cdot \vec{b} \)

The dot product of two vectors is calculated as:\[ \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z \]Substitute the components from previous steps:\[ \vec{a} \cdot \vec{b} = (0)(0.94) + (1.45)(0) + (2.85)(1.04) = 2.964 \]

Step 4: Calculate the Cross Product \( \vec{a} \times \vec{b} \)

The cross product is calculated using the determinant:\[ \vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 1.45 & 2.85 \ 0.94 & 0 & 1.04 \end{vmatrix} \]This expands to:\[ \mathbf{i}(1.45 \times 1.04 - 2.85 \times 0) - \mathbf{j}(0 \times 1.04 - 2.85 \times 0.94) + \mathbf{k}(0 \times 0 - 1.45 \times 0.94) \]Calculating each component, we get:\[ \vec{a} \times \vec{b} = \langle 1.508, -2.679, -1.363 \rangle \]

Step 5: Calculate the Magnitudes of \( \vec{a} \) and \( \vec{b} \)

The magnitude is calculated by:\[ |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = \sqrt{0^2 + 1.45^2 + 2.85^2} \approx 3.20 \]\[ |\vec{b}| = \sqrt{b_x^2 + b_y^2 + b_z^2} = \sqrt{0.94^2 + 0^2 + 1.04^2} \approx 1.40 \]

Step 6: Calculate the Angle Between \( \vec{a} \) and \( \vec{b} \)

The angle \( \theta \) between two vectors is given by:\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \]Substituting the values from previous steps:\[ \cos \theta = \frac{2.964}{3.20 \times 1.40} \approx 0.664 \]Thus, \[ \theta = \cos^{-1}(0.664) \approx 48.69^\circ \].

Key Concepts

Dot Product

The dot product is a way to multiply two vectors that results in a scalar, rather than another vector. It helps determine how much of one vector goes in the direction of another. The formula to find the dot product of vectors \( \vec{a} = \langle a_x, a_y, a_z \rangle \) and \( \vec{b} = \langle b_x, b_y, b_z \rangle \) is:

• \( \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z \)

This formula calculates a sum of products of their corresponding components. In this specific problem, since \( \vec{a} \) has only \( y \) and \( z \) components and \( \vec{b} \) has only \( x \) and \( z \) components, only the \( z \) components contribute:

• \( \vec{a} \cdot \vec{b} = 2.85 \times 1.04 = 2.964 \)

The result, a scalar value (2.964), implies that the vectors have some degree of alignment in the \( z \)-axis, illustrating the usefulness of the dot product in assessing vector similarity.

Cross Product

The cross product of two vectors results in a third vector that is perpendicular to the plane formed by the original vectors. For vectors \( \vec{a} = \langle a_x, a_y, a_z \rangle \) and \( \vec{b} = \langle b_x, b_y, b_z \rangle \), the cross product \( \vec{a} \times \vec{b} \) is determined by the determinant of a matrix involving the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \):

• \( \vec{a} \times \vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_x & a_y & a_z \ b_x & b_y & b_z \end{vmatrix} \)

The result is computed as:

• \( \mathbf{i}(a_y b_z - a_z b_y) - \mathbf{j}(a_x b_z - a_z b_x) + \mathbf{k}(a_x b_y - a_y b_x) \)

For this exercise, computing the cross product gives:

• \( \vec{a} \times \vec{b} = \langle 1.508, -2.679, -1.363 \rangle \)

The resulting vector is perpendicular to both \( \vec{a} \) and \( \vec{b} \), and its direction is determined by the right-hand rule. The magnitude of the cross product reflects the area of the parallelogram formed by \( \vec{a} \) and \( \vec{b} \).

Angle Between Vectors

The angle between two vectors is a valuable measure to understand how closely aligned they are within a plane. Typically, the cosine of the angle \( \theta \) between vectors \( \vec{a} \) and \( \vec{b} \) can be found using the dot product and the magnitudes (lengths) of the vectors:

• \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \)

To find the angle, follow these steps:

• Calculate the dot product \( \vec{a} \cdot \vec{b} \).
• Determine the magnitudes of each vector using \( |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \) and \( |\vec{b}| = \sqrt{b_x^2 + b_y^2 + b_z^2} \).
• Use the above formula to solve for \( \cos \theta \).

Applying these steps for this scenario:

• \( \cos \theta = \frac{2.964}{3.20 \times 1.40} \approx 0.664 \)

The angle is finally calculated using the inverse cosine:

• \( \theta = \cos^{-1}(0.664) \approx 48.69^\circ \)

This angle provides insight into the positioning of \( \vec{a} \) and \( \vec{b} \) in space, with a smaller angle indicating a more parallel alignment and a larger angle indicating more perpendicular vectors.