A force `F= 3hat(i) + 4hat(j) - 3hat(k)` is applied at the point P. whose position vector is `r= 2 hat(i) -2hat(j) - 3hat(k)`. What is the magnitude of the moment of the force about the origin?

Options

A. A. 23 units
B. B. 19 units
C. C. 18 units
D. D. 21 units

Answer

To solve the problem, we need to find the magnitude of the moment of the force about the origin. The moment of a force (also known as torque) is given by the cross product of the position vector and the force vector.

### Step-by-Step Solution:

1. **Identify the Given Vectors:**
- The force vector \( \mathbf{F} \) is given as:
\[ \mathbf{F} = 3\hat{i} + 4\hat{j} - 3\hat{k} \]
- The position vector \( \mathbf{r} \) is given as:
\[ \mathbf{r} = 2\hat{i} - 2\hat{j} - 3\hat{k} \]

2. **Calculate the Moment of the Force:**
- The moment \( \mathbf{M} \) about the origin is calculated using the cross product:
\[ \mathbf{M} = \mathbf{r} \times \mathbf{F} \]
- We can write this as:
\[ \mathbf{M} = (2\hat{i} - 2\hat{j} - 3\hat{k}) \times (3\hat{i} + 4\hat{j} - 3\hat{k}) \]

3. **Set Up the Cross Product Using Determinants:**
- We can express the cross product using a determinant:
\[ \mathbf{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -2 & -3 \\ 3 & 4 & -3 \end{vmatrix} \]

4. **Calculate the Determinant:**
- Expanding the determinant:
\[ \mathbf{M} = \hat{i} \begin{vmatrix} -2 & -3 \\ 4 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -3 \\ 3 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -2 \\ 3 & 4 \end{vmatrix} \]
- Calculating each of these 2x2 determinants:
- For \( \hat{i} \):
\[ (-2)(-3) - (-3)(4) = 6 + 12 = 18 \]
- For \( \hat{j} \):
\[ (2)(-3) - (-3)(3) = -6 + 9 = 3 \quad \text{(note the negative sign in front)} \]
- For \( \hat{k} \):
\[ (2)(4) - (-2)(3) = 8 + 6 = 14 \]
- Therefore, we have:
\[ \mathbf{M} = 18\hat{i} - 3\hat{j} + 14\hat{k} \]

5. **Find the Magnitude of the Moment:**
- The magnitude \( |\mathbf{M}| \) is calculated as:
\[ |\mathbf{M}| = \sqrt{(18)^2 + (-3)^2 + (14)^2} \]
- Calculating the squares:
\[ |\mathbf{M}| = \sqrt{324 + 9 + 196} = \sqrt{529} \]
- Thus, the magnitude is:
\[ |\mathbf{M}| = 23 \]

### Final Answer:
The magnitude of the moment of the force about the origin is \( 23 \) units.