A force of 3 units acts through the point `(4, -1, 7)` in the direction of the vector `9hat(i)+6hat(j)-2hat(k)`. Find the moment of the force about the point `(1, -3,2)` and the moment about the axes, parallel to the co - ordinate axes, which pass through `(1,-3, 2)`.

Answer

To solve the problem, we need to find the moment of a force about a point and then the moments about the axes parallel to the coordinate axes that pass through that point. Let's break it down step by step.

### Step 1: Define the Given Information
- The force \( F \) has a magnitude of 3 units and acts in the direction of the vector \( \vec{d} = 9\hat{i} + 6\hat{j} - 2\hat{k} \).
- The point through which the force acts is \( B(4, -1, 7) \).
- The point about which we want to find the moment is \( A(1, -3, 2) \).

### Step 2: Calculate the Unit Vector of the Force
To find the force vector \( \vec{F} \), we first need to calculate the unit vector in the direction of \( \vec{d} \).

1. Calculate the magnitude of \( \vec{d} \):
\[ |\vec{d}| = \sqrt{9^2 + 6^2 + (-2)^2} = \sqrt{81 + 36 + 4} = \sqrt{121} = 11 \]

2. The unit vector \( \hat{d} \) in the direction of \( \vec{d} \) is:
\[ \hat{d} = \frac{\vec{d}}{|\vec{d}|} = \frac{9\hat{i} + 6\hat{j} - 2\hat{k}}{11} \]

3. Now, calculate the force vector \( \vec{F} \):
\[ \vec{F} = 3 \hat{d} = 3 \left(\frac{9\hat{i} + 6\hat{j} - 2\hat{k}}{11}\right) = \frac{27}{11}\hat{i} + \frac{18}{11}\hat{j} - \frac{6}{11}\hat{k} \]

### Step 3: Calculate the Position Vector \( \vec{r_{AB}} \)
The position vector \( \vec{r_{AB}} \) from point \( A \) to point \( B \) is given by:
\[ \vec{r_{AB}} = \vec{OB} - \vec{OA} = (4\hat{i} - 1\hat{j} + 7\hat{k}) - (1\hat{i} - 3\hat{j} + 2\hat{k}) = (4-1)\hat{i} + (-1 + 3)\hat{j} + (7-2)\hat{k} = 3\hat{i} + 2\hat{j} + 5\hat{k} \]

### Step 4: Calculate the Moment of the Force about Point A
The moment \( \vec{M_A} \) about point \( A \) is given by the cross product:
\[ \vec{M_A} = \vec{r_{AB}} \times \vec{F} \]

1. Set up the determinant for the cross product:
\[ \vec{M_A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 5 \\ \frac{27}{11} & \frac{18}{11} & -\frac{6}{11} \end{vmatrix} \]

2. Calculate the determinant:
\[ \vec{M_A} = \hat{i} \left(2 \cdot -\frac{6}{11} - 5 \cdot \frac{18}{11}\right) - \hat{j} \left(3 \cdot -\frac{6}{11} - 5 \cdot \frac{27}{11}\right) + \hat{k} \left(3 \cdot \frac{18}{11} - 2 \cdot \frac{27}{11}\right) \]

3. Simplifying each component:
- \( \hat{i} \) component:
\[ 2 \cdot -\frac{6}{11} - 5 \cdot \frac{18}{11} = -\frac{12}{11} - \frac{90}{11} = -\frac{102}{11} \]
- \( \hat{j} \) component:
\[ 3 \cdot -\frac{6}{11} - 5 \cdot \frac{27}{11} = -\frac{18}{11} - \frac{135}{11} = -\frac{153}{11} \]
- \( \hat{k} \) component:
\[ 3 \cdot \frac{18}{11} - 2 \cdot \frac{27}{11} = \frac{54}{11} - \frac{54}{11} = 0 \]

Thus, the moment vector is:
\[ \vec{M_A} = -\frac{102}{11}\hat{i} + \frac{153}{11}\hat{j} + 0\hat{k} \]

### Step 5: Moments about the Axes
The moments about the axes parallel to the coordinate axes that pass through point \( A \) can be calculated as follows:

1. **Moment about the x-axis**:
\[ M_x = -\frac{102}{11} \]

2. **Moment about the y-axis**:
\[ M_y = \frac{153}{11} \]

3. **Moment about the z-axis**:
\[ M_z = 0 \]

### Final Answer
The moment of the force about point \( A(1, -3, 2) \) is:
\[ \vec{M_A} = -\frac{102}{11}\hat{i} + \frac{153}{11}\hat{j} + 0\hat{k} \]
The moments about the axes are:
- Moment about x-axis: \( -\frac{102}{11} \)
- Moment about y-axis: \( \frac{153}{11} \)
- Moment about z-axis: \( 0 \)