To solve the problem of finding the dot product \( \mathbf{a} \cdot \mathbf{b} \) and the cross product \( \mathbf{a} \times \mathbf{b} \) of the two vectors given their orientations and magnitudes, we can break it down into manageable steps. Let's start by determining the components of each vector based on the information provided.
Defining Vector Components
We know that:
- Vector \( \mathbf{a} \) lies in the yz-plane, making its x-component zero.
- Vector \( \mathbf{b} \) lies in the xz-plane, making its y-component zero.
Calculating Vector \( \mathbf{a} \)
Given that vector \( \mathbf{a} \) has a magnitude of 3.20 units and is at an angle of 63.0 degrees from the +y-axis, we can find its components as follows:
- The y-component is given by \( a_y = 3.20 \cdot \cos(63.0^\circ) \).
- The z-component is given by \( a_z = 3.20 \cdot \sin(63.0^\circ) \).
Calculating these values:
- Using \( \cos(63.0^\circ) \approx 0.454 \) and \( \sin(63.0^\circ) \approx 0.891 \):
- \( a_y \approx 3.20 \cdot 0.454 \approx 1.45 \) units
- \( a_z \approx 3.20 \cdot 0.891 \approx 2.85 \) units
Thus, vector \( \mathbf{a} \) can be expressed as:
Vector \( \mathbf{a} = (0, 1.45, 2.85) \)
Calculating Vector \( \mathbf{b} \)
For vector \( \mathbf{b} \), which has a magnitude of 1.40 units and is at an angle of 48 degrees from the +x-axis, we find its components similarly:
- The x-component is \( b_x = 1.40 \cdot \cos(48.0^\circ) \).
- The z-component is \( b_z = 1.40 \cdot \sin(48.0^\circ) \).
Calculating these values:
- Using \( \cos(48.0^\circ) \approx 0.669 \) and \( \sin(48.0^\circ) \approx 0.743 \):
- \( b_x \approx 1.40 \cdot 0.669 \approx 0.94 \) units
- \( b_z \approx 1.40 \cdot 0.743 \approx 1.04 \) units
Thus, vector \( \mathbf{b} \) can be expressed as:
Vector \( \mathbf{b} = (0.94, 0, 1.04) \)
Finding the Dot Product \( \mathbf{a} \cdot \mathbf{b} \)
The dot product of two vectors is calculated using the formula:
\( \mathbf{a} \cdot \mathbf{b} = a_x b_x + a_y b_y + a_z b_z \)
Substituting the components we found:
- \( a_x = 0 \), \( b_x = 0.94 \)
- \( a_y = 1.45 \), \( b_y = 0 \)
- \( a_z = 2.85 \), \( b_z = 1.04 \)
Calculating the dot product:
\( \mathbf{a} \cdot \mathbf{b} = (0)(0.94) + (1.45)(0) + (2.85)(1.04) = 0 + 0 + 2.964 = 2.964 \)
Calculating the Cross Product \( \mathbf{a} \times \mathbf{b} \)
The cross product is determined using the determinant of a matrix formed by the unit vectors and the components of the vectors:
\( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1.45 & 2.85 \\ 0.94 & 0 & 1.04 \end{vmatrix} \)
Calculating this determinant, we can expand it:
- For \( \mathbf{i} \): \( \mathbf{i}(1.45 \cdot 1.04 - 0 \cdot 2.85) = 1.508 \mathbf{i} \)
- For \( \mathbf{j} \): \( -\mathbf{j}(0 \cdot 1.04 - 0.94 \cdot 2.85) = 2.679 \mathbf{j} \)
- For \( \mathbf{k} \): \( \mathbf{k}(0 \cdot 0 - 0.94 \cdot 1.45) = -1.363 \mathbf{k} \)
Combining these results, we find:
\( \mathbf{a} \times \mathbf{b} = (1.508, -2.679, -1.363) \)
Final Results
In summary, we have:
- Dot Product \( \mathbf{a} \cdot \mathbf{b} \approx 2.964 \)
- Cross Product \( \mathbf{a} \times \mathbf{b} = (1.508, -2.679, -1.363) \)
This approach illustrates how to break down vector problems into their components and apply vector operations systematically. If you have
