Problem 40 - Fundamentals of Physics

Question

Displacement \(\vec{d}_{1}\) is in the \(y z\) plane \(63.0^{\circ}\) from the positive direction of the \(y\) axis, has a positive \(z\) component, and has a magnitude of \(4.50 \mathrm{~m}\). Displacement \(\vec{d}_{2}\) is in the \(x z\) plane \(30.0^{\circ}\) from the positive direction of the \(x\) axis, has a positive \(z\) component, and has magnitude \(1.40 \mathrm{~m}\). What are (a) \(\vec{d}_{1} \cdot \vec{d}_{2}\), (b) \(\vec{d}_{1} \times \vec{d}_{2}\), and (c) the angle between \(\vec{d}_{1}\) and \(\vec{d}_{2}\) ?

Step-by-Step Solution

Answer

(a) \( \vec{d}_{1} \cdot \vec{d}_{2} = 2.804 \) m, (b) \( \vec{d}_{1} \times \vec{d}_{2} = (1.419, 4.8552, -2.457) \) m^2, (c) angle \( \approx 63.53^{\circ} \).

Step 1: Find Components of \( \vec{d}_{1} \)

Given that \( \vec{d}_{1} \) is in the \( yz \) plane, it only has \( y \) and \( z \) components. We use the angle \( \theta = 63.0^{\circ} \) measured from the positive direction of the \( y \) axis. Thus: \( d_{1y} = 4.50 \cos(63^{\circ}) = 2.027 \, \text{m} \), and \( d_{1z} = 4.50 \sin(63^{\circ}) = 4.006 \, \text{m} \). Therefore, \( \vec{d}_{1} = (0, 2.027, 4.006) \).

Step 2: Find Components of \( \vec{d}_{2} \)

Since \( \vec{d}_{2} \) is in the \( xz \) plane, it has \( x \) and \( z \) components. From the angle \( 30.0^{\circ} \) with the \( x \) axis, we have: \( d_{2x} = 1.40 \cos(30^{\circ}) = 1.212 \, \text{m} \), and \( d_{2z} = 1.40 \sin(30^{\circ}) = 0.700 \, \text{m} \). Thus, \( \vec{d}_{2} = (1.212, 0, 0.700) \).

Step 3: Calculate Dot Product \( \vec{d}_{1} \cdot \vec{d}_{2} \)

The dot product is calculated using: \( \vec{d}_{1} \cdot \vec{d}_{2} = d_{1x}d_{2x} + d_{1y}d_{2y} + d_{1z}d_{2z} \). Substituting the values, we get: \( 0 \times 1.212 + 2.027 \times 0 + 4.006 \times 0.700 = 2.804 \).

Step 4: Calculate Cross Product \( \vec{d}_{1} \times \vec{d}_{2} \)

To find the cross product, use the determinant formula: \[ \vec{d}_{1} \times \vec{d}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2.027 & 4.006 \ 1.212 & 0 & 0.700 \end{vmatrix} \]. Compute the individual components: \( \hat{i} = (2.027)(0.700) - (4.006)(0) = 1.419 \, \hat{j} = (4.006)(1.212) - (0)(0.700) = 4.8552 \, and \hat{k} = (0)(0) - (2.027)(1.212) = -2.457 \). The cross product is \( \vec{d}_{1} \times \vec{d}_{2} = (1.419, 4.8552, -2.457) \) \text{m}^2.

Step 5: Calculate Angle Between \( \vec{d}_{1} \) and \( \vec{d}_{2} \)

The angle \( \theta \) between the vectors can be found using the formula: \( \cos \theta = \frac{\vec{d}_{1} \cdot \vec{d}_{2}}{\|\vec{d}_{1}\| \|\vec{d}_{2}\|} \). Compute magnitudes: \( \|\vec{d}_{1}\| = 4.50 \) \, \( \|\vec{d}_{2}\| = 1.40 \). Then, \( \cos \theta = \frac{2.804}{4.50 \times 1.40} = 0.446 \). Thus, \( \theta = \cos^{-1}(0.446) \approx 63.53^{\circ} \).

Key Concepts

Dot Product

The dot product, also known as the scalar product, is a key concept in vector calculus. It provides a measure of how much one vector extends in the direction of another. To calculate the dot product of two vectors, you multiply their corresponding components and sum the results. Consider two vectors, \( \vec{A} = (a_x, a_y, a_z) \) and \( \vec{B} = (b_x, b_y, b_z) \). The dot product is given by:

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

This operation results in a scalar, not a vector. Therefore, the dot product only tells us about magnitudes and directions relative to each other. If the dot product is zero, the vectors are perpendicular. In the original problem, the dot product of \( \vec{d}_1 \) and \( \vec{d}_2 \) is \( 2.804 \), showing a certain degree of alignment between them.

Cross Product

The cross product, or vector product, is another fundamental concept in vector calculus. This operation produces a vector that is perpendicular to the plane formed by the initial two vectors. If two vectors \( \vec{A} = (a_x, a_y, a_z) \) and \( \vec{B} = (b_x, b_y, b_z) \) are given, the cross product \( \vec{A} \times \vec{B} \) can be computed using the determinant:

• \( \vec{A} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \ a_x & a_y & a_z \ b_x & b_y & b_z\end{vmatrix} \)

This expression expands to find each component:

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

The result is a vector perpendicular to both \( \vec{A} \) and \( \vec{B} \). In the given problem, the cross product produces a vector \( (1.419, 4.8552, -2.457) \) \(\text{m}^2\), which is orthogonal to \( \vec{d}_1 \) and \( \vec{d}_2 \).

Angle Between Vectors

Understanding the angle between vectors is important for determining how they relate spatially. The angle \( \theta \) can be determined using the dot product formula, combined with the magnitudes of vectors.

• Given by the expression: \( \cos \theta = \frac{\vec{A} \cdot \vec{B}}{\|\vec{A}\| \|\vec{B}\|} \)

Here, \( \|\vec{A}\| \) and \( \|\vec{B}\| \) are the magnitudes (lengths) of the vectors, computed using:

• \( \|\vec{A}\| = \sqrt{a_x^2 + a_y^2 + a_z^2} \)

Solving for the angle requires using the inverse cosine function (\( \cos^{-1} \)). When the dot product is positive, the angle is acute, meaning it's less than \( 90^\circ \), and when negative, the angle is obtuse, greater than \( 90^\circ \). In the example problem, the calculated angle is approximately \( 63.53^{\circ} \), showing a modest divergence between \( \vec{d}_1 \) and \( \vec{d}_2 \).