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 solve the problem, we need to find the moment of a force about a point. Here is the step-by-step solution:

### Step 1: Identify the Given Data
We have a force of magnitude \( F = 6 \) acting along the vector \( \mathbf{v} = (9, 6, -2) \). The force passes through the point \( A(4, -1, -7) \) and we need to find the moment about the point \( O(1, -3, 2) \).

### Step 2: Calculate the Unit Vector of the Force
The unit vector \( \mathbf{u} \) in the direction of the vector \( \mathbf{v} \) is calculated as follows:
\[ \mathbf{u} = \frac{\mathbf{v}}{|\mathbf{v}|} \]
First, we find the magnitude \( |\mathbf{v}| \):
\[ |\mathbf{v}| = \sqrt{9^2 + 6^2 + (-2)^2} = \sqrt{81 + 36 + 4} = \sqrt{121} = 11 \]
Thus, the unit vector is:
\[ \mathbf{u} = \frac{(9, 6, -2)}{11} = \left(\frac{9}{11}, \frac{6}{11}, -\frac{2}{11}\right) \]

### Step 3: Calculate the Force Vector
Now, we can find the force vector \( \mathbf{F} \):
\[ \mathbf{F} = F \cdot \mathbf{u} = 6 \cdot \left(\frac{9}{11}, \frac{6}{11}, -\frac{2}{11}\right) = \left(\frac{54}{11}, \frac{36}{11}, -\frac{12}{11}\right) \]

### Step 4: Calculate the Position Vector \( \mathbf{R} \)
The position vector \( \mathbf{R} \) from point \( O(1, -3, 2) \) to point \( A(4, -1, -7) \) is given by:
\[ \mathbf{R} = A - O = (4 - 1, -1 + 3, -7 - 2) = (3, 2, -9) \]

### Step 5: Calculate the Moment of the Force
The moment \( \mathbf{M} \) of the force about point \( O \) is given by the cross product:
\[ \mathbf{M} = \mathbf{R} \times \mathbf{F} \]
We can set up the determinant to calculate the cross product:
\[ \mathbf{M} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & -9 \\ \frac{54}{11} & \frac{36}{11} & -\frac{12}{11} \end{vmatrix} \]

### Step 6: Calculate the Determinant
Calculating the determinant, we have:
\[ \mathbf{M} = \mathbf{i} \left(2 \cdot -\frac{12}{11} - (-9) \cdot \frac{36}{11}\right) - \mathbf{j} \left(3 \cdot -\frac{12}{11} - (-9) \cdot \frac{54}{11}\right) + \mathbf{k} \left(3 \cdot \frac{36}{11} - 2 \cdot \frac{54}{11}\right) \]
Calculating each component:
1. For \( \mathbf{i} \):
\[ 2 \cdot -\frac{12}{11} + 9 \cdot \frac{36}{11} = -\frac{24}{11} + \frac{324}{11} = \frac{300}{11} \]
2. For \( \mathbf{j} \):
\[ 3 \cdot -\frac{12}{11} + 9 \cdot \frac{54}{11} = -\frac{36}{11} + \frac{486}{11} = \frac{450}{11} \]
3. For \( \mathbf{k} \):
\[ 3 \cdot \frac{36}{11} - 2 \cdot \frac{54}{11} = \frac{108}{11} - \frac{108}{11} = 0 \]

Thus, the moment vector is:
\[ \mathbf{M} = \left(\frac{300}{11}, -\frac{450}{11}, 0\right) \]

### Final Answer
The moment of the force about the point \( O(1, -3, 2) \) is:
\[ \mathbf{M} = \left(\frac{300}{11}, -\frac{450}{11}, 0\right) \]