Question

Consider \( \mathbf{C} = \mathbf{A} \times \mathbf{B} \), where \( \mathbf{A} = 3\hat{i} - 2\hat{j} + 5\hat{k} \) and \( \mathbf{B} \) is a unit vector in the \( xy \)-plane, making an angle of 37° with the \( x \)-axis. The projection of \( \mathbf{C} \) on the \( x \)-axis is ................ (round off to one decimal place)

Hint:

For cross products, remember to use the determinant form, and for the projection on an axis, use the corresponding component of the result vector.

Answer 1

We are given: \[ \mathbf{A} = 3\hat{i} - 2\hat{j} + 5\hat{k}, \quad \mathbf{B} = \cos(37^\circ) \hat{i} + \sin(37^\circ) \hat{j} \] since \( \mathbf{B} \) is a unit vector in the \( xy \)-plane and makes an angle of \( 37^\circ \) with the \( x \)-axis.
Step 1: Compute the cross product \( \mathbf{A} \times \mathbf{B} \).
Using the determinant form for the cross product: \[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -2 & 5 \\ \cos(37^\circ) & \sin(37^\circ) & 0 \end{vmatrix} \] Expanding this determinant: \[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} -2 & 5 \\ \sin(37^\circ) & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 5 \\ \cos(37^\circ) & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -2 \\ \cos(37^\circ) & \sin(37^\circ) \end{vmatrix} \] Solving the determinants: \[ \mathbf{A} \times \mathbf{B} = \hat{i}(0 - 5\sin(37^\circ)) - \hat{j}(0 - 5\cos(37^\circ)) + \hat{k}(3\sin(37^\circ) + 2\cos(37^\circ)) \] \[ \mathbf{A} \times \mathbf{B} = -5\sin(37^\circ)\hat{i} + 5\cos(37^\circ)\hat{j} + \big(3\sin(37^\circ) + 2\cos(37^\circ)\big)\hat{k} \] Step 2: Find the projection of \( \mathbf{C} \) on the \( x \)-axis.
The projection of \( \mathbf{C} = \mathbf{A} \times \mathbf{B} \) on the \( x \)-axis is the \( x \)-component: \[ \text{Projection on } x\text{-axis} = -5\sin(37^\circ) \] Using \( \sin(37^\circ) \approx 0.6018 \): \[ \text{Projection on } x\text{-axis} = -5(0.6018) \approx -3.009 \] Final Answer: \[ \boxed{-3.0} \]