Question

If \( \mathbf{F} = (4\hat{i} - 10\hat{j}) \, \text{N} \) and \( \mathbf{r} = (-5\hat{i} - 3\hat{j}) \, \text{m} \), then \( \mathbf{r} \times \mathbf{F} \) is:

• A. \( (-20\hat{i} + 3\hat{j}) \, \text{Nm} \)
• B. \( 62 \, \text{kNm} \)
• C. \( \frac{10}{\sqrt{13}} \, \text{Nm} \)
• D. \( 38 \, \text{Nm} \)

Hint:

For cross products, remember that the result is a vector perpendicular to both \( \mathbf{r} \) and \( \mathbf{F} \), and its magnitude is given by \( |\mathbf{r} \times \mathbf{F}| = r F \sin \theta \), where \( \theta \) is the angle between the two vectors.

Answer 1

The Correct Option is B

The cross product of two vectors \( \mathbf{r} \) and \( \mathbf{F} \) is given by the formula: \[ \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ r_x & r_y & r_z \\ F_x & F_y & F_z \end{vmatrix} \] Given: \[ \mathbf{r} = (-5\hat{i} - 3\hat{j}) \, \text{m}, \quad \mathbf{F} = (4\hat{i} - 10\hat{j}) \, \text{N} \] Substitute into the determinant formula: \[ \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & -3 & 0 \\ 4 & -10 & 0 \end{vmatrix} \] Calculating the determinant: \[ = \hat{i} \begin{vmatrix} -3 & 0 \\ -10 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} -5 & 0 \\ 4 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} -5 & -3 \\ 4 & -10 \end{vmatrix} \] \[ = \hat{i} (0 - 0) - \hat{j} (0 - 0) + \hat{k} \left[(-5)(-10) - (4)(-3)\right] \] \[ = 0 \hat{i} - 0 \hat{j} + \hat{k} (50 + 12) = 62 \hat{k} \, \text{Nm} \] Thus, the correct answer is \( \mathbf{62 \, \text{kNm}} \), and the correct option is (2).

Answer 2

Step 1: Write down the given vectors
\(\mathbf{F} = 4\hat{i} - 10\hat{j} \, \text{N}\)
\(\mathbf{r} = -5\hat{i} - 3\hat{j} \, \text{m}\)

Step 2: Recall the cross product formula for two vectors in \( \hat{i}, \hat{j}, \hat{k} \) components
\[ \mathbf{r} \times \mathbf{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ r_x & r_y & r_z \\ F_x & F_y & F_z \\ \end{vmatrix} \]
Since both vectors have no \(\hat{k}\) component, \( r_z = 0 \), \( F_z = 0 \)

Step 3: Substitute values
\[ \mathbf{r} \times \mathbf{F} = \hat{i} \begin{vmatrix} -3 & 0 \\ -10 & 0 \end{vmatrix} - \hat{j} \begin{vmatrix} -5 & 0 \\ 4 & 0 \end{vmatrix} + \hat{k} \begin{vmatrix} -5 & -3 \\ 4 & -10 \end{vmatrix} \]
The determinants with zeros in the last column are zero:
\[ = 0 \hat{i} - 0 \hat{j} + \hat{k}((-5)(-10) - (-3)(4)) \]

Step 4: Calculate the determinant for \(\hat{k}\)
\[ = \hat{k} (50 + 12) = 62 \hat{k} \]

Final answer: \( \mathbf{r} \times \mathbf{F} = 62 \hat{k} \, \text{Nm} \)