Question:

Vector Components The given vectors are converted into their Cartesian components using the formulas \(x = r \cos \theta\) and \(y = r \sin \theta\). 1. Vector \(A\): \(A_x = 1 \cos(30^\circ) = \frac{\sqrt{3}}{2}\), \(A_y = 1 \sin(30^\circ) = \frac{1}{2}\). 2. Vector \(B\): \(B_x = 2 \cos(60^\circ) = 1\), \(B_y = 2 \sin(60^\circ) = \sqrt{3}\). 3. Vector \(C\): \(C_x = 1.5 \cos(180^\circ) = -1.5\), \(C_y = 1.5 \sin(180^\circ) = 0\). 4. Vector \(D\): \(D_x = 3 \cos(225^\circ) = -\frac{3\sqrt{2}}{2}\), \(D_y = 3 \sin(225^\circ) = -\frac{3\sqrt{2}}{2}\). Vector Addition and Subtraction The resultant vector \(R = A+B+C-D\) is found by adding and subtracting the corresponding components. 1. The x-component of the resultant vector is calculated as \(R_x = A_x + B_x + C_x - D_x = \frac{\sqrt{3}}{2} + 1 - 1.5 - (-\frac{3\sqrt{2}}{2}) \approx 0.866 + 1 - 1.5 + 2.121 = 2.487\). 2. The y-component of the resultant vector is calculated as \(R_y = A_y + B_y + C_y - D_y = \frac{1}{2} + \sqrt{3} + 0 - (-\frac{3\sqrt{2}}{2}) \approx 0.5 + 1.732 + 2.121 = 4.353\). \[\begin{array}{|c|c|c|} \hline \text{3. The length (magnitude) of the resultant vector is found using the Pythagorean theorem:} & \text{R} & \text{= \sqrt{R_x^2 + R_y^2} = \sqrt{(2.487)^2 + (4.353)^2} \approx \sqrt{6.185 + 18.949} = \sqrt{25.134} \approx 5.013.} \\ \hline \end{array}\] 4. The angle of the resultant vector is found using the arctangent function: \(\theta_R = \arctan(\frac{R_y}{R_x}) = \arctan(\frac{4.353}{2.487}) \approx \arctan(1.750) \approx 60.3^\circ\). Dot Product \[\begin{array}{|c|c|c|c|} \hline \text{The dot product of vectors A$ and $B$ is calculated using the formula $A \cdot B =} & \text{A} & \text{B} & \text{\cos \phi where \(\phi\) is the angle between the vectors.} \\ \hline \end{array}\] 1. The angle between \(A\) and \(B\) is \(\phi = 60^\circ - 30^\circ = 30^\circ\). 2. The dot product is calculated as \(A \cdot B = (1)(2) \cos(30^\circ) = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.732\). Cross Product \[\begin{array}{|c|c|c|c|c|c|} \hline \text{The cross product of two planar vectors C$ and $D$ is calculated using the formula $} & \text{C \times D} & \text{=} & \text{C} & \text{D} & \text{\sin \phi where \(\phi\) is the angle between the vectors. The direction is determined by the right-hand rule.} \\ \hline \end{array}\] 1. The angle between \(C\) and \(D\) is \(\phi = 225^\circ - 180^\circ = 45^\circ\). \[\begin{array}{|c|c|c|} \hline \text{2. The length of the cross product is calculated as} & \text{C \times D} & \text{= (1.5)(3) \sin(45^\circ) = 4.5 \cdot \frac{\sqrt{2}}{2} \approx 3.182.} \\ \hline \end{array}\] 3. The direction of the cross product is perpendicular to the plane containing \(C\) and \(D\). Since \(D\) is counterclockwise from \(C\), the direction of \(C \times D\) is out of the page (positive z-direction). Final Answer The final answers are: • For \(A+B+C-D\): The length of the resulting vector is approximately $5.013$, and its angle is approximately \(60.3^\circ\). • For \(A \cdot B\): The dot product is approximately $1.732$. • For \(C \times D\): The length of the resulting vector is approximately $3.182$, and its direction is out of the page (positive z-direction). AI responses may include mistakes.