Let `vec(F)=2hat(i)+4hat(j)+3hat(k)` at the point P with position vector `hat(i)-hat(j)+3hat(k)`. Find the moment of `vec(F)` about the line through the origin O in the direction of the vector `vec(a)=hat(i)+2hat(j)+2hat(k)`.

Answer

To find the moment of the vector \(\vec{F} = 2\hat{i} + 4\hat{j} + 3\hat{k}\) about the line through the origin in the direction of the vector \(\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}\), we can follow these steps:

### Step 1: Determine the Position Vector
The position vector at point \(P\) is given as:
\[ \vec{r_P} = \hat{i} - \hat{j} + 3\hat{k} \]

### Step 2: Calculate the Moment about Point \(P\)
The moment \(\vec{M}\) about point \(P\) is calculated using the formula:
\[ \vec{M} = \vec{r_P} \times \vec{F} \]
Where \(\vec{F} = 2\hat{i} + 4\hat{j} + 3\hat{k}\).

Set up the cross product:
\[ \vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & 4 & 3 \end{vmatrix} \]

### Step 3: Calculate the Determinant
Calculating the determinant:
\[ \vec{M} = \hat{i} \begin{vmatrix} -1 & 3 \\ 4 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ 2 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 2 & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
- For \(\hat{i}\):
\[ (-1)(3) - (3)(4) = -3 - 12 = -15 \]
- For \(\hat{j}\):
\[ (1)(3) - (3)(2) = 3 - 6 = -3 \]
- For \(\hat{k}\):
\[ (1)(4) - (-1)(2) = 4 + 2 = 6 \]

### Step 4: Combine the Results
Thus, the moment vector \(\vec{M}\) is:
\[ \vec{M} = -15\hat{i} + 3\hat{j} + 6\hat{k} \]

### Step 5: Find the Unit Vector in the Direction of Line \(\vec{a}\)
The unit vector \(\hat{a}\) in the direction of \(\vec{a}\) is given by:
\[ \hat{a} = \frac{\vec{a}}{|\vec{a}|} \]
Where:
\[ |\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
Thus,
\[ \hat{a} = \frac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k}) \]

### Step 6: Calculate the Moment about the Line
The moment about the line is given by:
\[ M_L = \frac{\vec{M} \cdot \hat{a}}{|\hat{a}|} \]
Since \(|\hat{a}| = 1\), we need to calculate \(\vec{M} \cdot \hat{a}\):
\[ \vec{M} \cdot \hat{a} = (-15\hat{i} + 3\hat{j} + 6\hat{k}) \cdot \left(\frac{1}{3}(\hat{i} + 2\hat{j} + 2\hat{k})\right) \]
Calculating the dot product:
\[ = \frac{1}{3} \left(-15 \cdot 1 + 3 \cdot 2 + 6 \cdot 2\right) = \frac{1}{3} \left(-15 + 6 + 12\right) = \frac{1}{3} \left{3\right} = 1 \]

### Final Answer
Thus, the moment of \(\vec{F}\) about the line is:
\[ \boxed{1} \]