A force of magnitude 6 acts along the vector `(9, 6, -2)` and passes through a point `A(4, -1,-7)`. Then moment of force about the point `O(1, -3, 2)` is

Options

A. `(150)/(11)(2hat(i)-3hat(j))`
B. `(6)/(11)(50hat(i)-75hat(j)+36hat(k))`
C. `150(2hat(i)-3hat(k))`
D. `6(50hat(i)-75hat(j)+36hat(k))`

Answer

To find the moment of the force about the point O, we will follow these steps:

### Step 1: Find the unit vector along the direction of the force
The force acts along the vector \( \vec{F} = (9, 6, -2) \). The magnitude of this vector is calculated as follows:

\[ |\vec{F}| = \sqrt{9^2 + 6^2 + (-2)^2} = \sqrt{81 + 36 + 4} = \sqrt{121} = 11 \]

Now, the unit vector \( \hat{F} \) in the direction of the force is:

\[ \hat{F} = \frac{\vec{F}}{|\vec{F}|} = \frac{(9, 6, -2)}{11} = \left( \frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \right) \]

### Step 2: Find the force vector \( \vec{F} \)
The force vector \( \vec{F} \) with a magnitude of 6 is:

\[ \vec{F} = 6 \hat{F} = 6 \left( \frac{9}{11}, \frac{6}{11}, -\frac{2}{11} \right) = \left( \frac{54}{11}, \frac{36}{11}, -\frac{12}{11} \right) \]

### Step 3: Find the position vector \( \vec{r} \)
The position vector \( \vec{r} \) from point O(1, -3, 2) to point A(4, -1, -7) is calculated as follows:

\[ \vec{r} = \vec{A} - \vec{O} = (4, -1, -7) - (1, -3, 2) = (4-1, -1+3, -7-2) = (3, 2, -9) \]

### Step 4: Calculate the moment of the force about point O
The moment \( \vec{M} \) of the force about point O is given by the cross product:

\[ \vec{M} = \vec{r} \times \vec{F} \]

Using the determinant form for the cross product:

\[ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & -9 \\ \frac{54}{11} & \frac{36}{11} & -\frac{12}{11} \end{vmatrix} \]

### Step 5: Compute the determinant
Calculating the determinant, we expand it as follows:

\[ \vec{M} = \hat{i} \begin{vmatrix} 2 & -9 \\ \frac{36}{11} & -\frac{12}{11} \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & -9 \\ \frac{54}{11} & -\frac{12}{11} \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 2 \\ \frac{54}{11} & \frac{36}{11} \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ 2 \cdot \left(-\frac{12}{11}\right) - (-9) \cdot \frac{36}{11} = -\frac{24}{11} + \frac{324}{11} = \frac{300}{11} \]

2. For \( \hat{j} \):
\[ 3 \cdot \left(-\frac{12}{11}\right) - (-9) \cdot \frac{54}{11} = -\frac{36}{11} + \frac{486}{11} = \frac{450}{11} \]

3. For \( \hat{k} \):
\[ 3 \cdot \frac{36}{11} - 2 \cdot \frac{54}{11} = \frac{108}{11} - \frac{108}{11} = 0 \]

### Step 6: Combine the results
Thus, the moment vector is:

\[ \vec{M} = \frac{300}{11} \hat{i} - \frac{450}{11} \hat{j} + 0 \hat{k} \]

### Step 7: Final result
The moment of the force about point O is:

\[ \vec{M} = \left( \frac{300}{11}, -\frac{450}{11}, 0 \right) \]