Questions: Problem 8: (11% of Assignment Value) In the diagram, the dashed lines are parallel to the x axis. The magnitudes of the vectors are A=10.5, B=6.6, D=23, and F=23. The angles, in degrees, are θA=30.0, θB=53.0, θD=37.0, and θF=30.0. - Part (a) v What is (A × F) · D ? (A × F) · D=0.000 Correct! Part (b) What is (A × D) ·(D × B) ? (A × D) ·(D × B)=

Solution Steps

We need to find the value of \((\mathbf{A} \times \mathbf{D}) \cdot (\mathbf{D} \times \mathbf{B})\). This involves calculating the cross products of vectors \(\mathbf{A}\) and \(\mathbf{D}\), and \(\mathbf{D}\) and \(\mathbf{B}\), and then taking the dot product of the resulting vectors.

The cross product of two vectors \(\mathbf{U}\) and \(\mathbf{V}\) is given by: \[ \mathbf{U} \times \mathbf{V} = |\mathbf{U}||\mathbf{V}|\sin(\theta)\hat{n} \] where \(\theta\) is the angle between the vectors and \(\hat{n}\) is the unit vector perpendicular to the plane containing \(\mathbf{U}\) and \(\mathbf{V}\).

For \(\mathbf{A} \times \mathbf{D}\): \[ |\mathbf{A} \times \mathbf{D}| = |A||D|\sin(\theta_A - \theta_D) \] Given \(A = 10.5\), \(D = 23\), \(\theta_A = 30.0^\circ\), and \(\theta_D = 37.0^\circ\): \[ |\mathbf{A} \times \mathbf{D}| = 10.5 \times 23 \times \sin(30.0^\circ - 37.0^\circ) = 10.5 \times 23 \times \sin(-7.0^\circ) \] \[ \sin(-7.0^\circ) = -\sin(7.0^\circ) \approx -0.1219 \] \[ |\mathbf{A} \times \mathbf{D}| = 10.5 \times 23 \times (-0.1219) \approx -29.4872 \]

For \(\mathbf{D} \times \mathbf{B}\): \[ |\mathbf{D} \times \mathbf{B}| = |D||B|\sin(\theta_D - \theta_B) \] Given \(D = 23\), \(B = 6.6\), \(\theta_D = 37.0^\circ\), and \(\theta_B = 53.0^\circ\): \[ |\mathbf{D} \times \mathbf{B}| = 23 \times 6.6 \times \sin(37.0^\circ - 53.0^\circ) = 23 \times 6.6 \times \sin(-16.0^\circ) \] \[ \sin(-16.0^\circ) = -\sin(16.0^\circ) \approx -0.2756 \] \[ |\mathbf{D} \times \mathbf{B}| = 23 \times 6.6 \times (-0.2756) \approx -41.7487 \]

The dot product of two vectors \(\mathbf{U}\) and \(\mathbf{V}\) is given by: \[ \mathbf{U} \cdot \mathbf{V} = |\mathbf{U}||\mathbf{V}|\cos(\theta) \] Since the cross products \(\mathbf{A} \times \mathbf{D}\) and \(\mathbf{D} \times \mathbf{B}\) are perpendicular to the plane containing the original vectors, their dot product is zero: \[ (\mathbf{A} \times \mathbf{D}) \cdot (\mathbf{D} \times \mathbf{B}) = 0 \]

Final Answer