The velocity of particle is `3hat(i) +2hat(j)+3hat(k)`. Find the vector component of the velocity along the line `hat(i) -hat(j)+hat(k)`.

Options

A. `(4)/(3)(hat(i)-hat(j)+hat(k))`
B. `(2)/(3)(hat(i)+hat(j)+hat(k))`
C. `(2)/(3)(3hat(i)+2hat(j)+3hat(k))`
D. None

Explanation

To find the vector component of the velocity along the line defined by the vector \(\hat{i} - \hat{j} + \hat{k}\), we can follow these steps:

### Step 1: Identify the vectors
Let the velocity vector \( \vec{v} = 3\hat{i} + 2\hat{j} + 3\hat{k} \) and the direction vector along the line be \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \).

### Step 2: Calculate the magnitude of vector \( \vec{b} \)
The magnitude of vector \( \vec{b} \) is calculated using the formula:
\[ |\vec{b}| = \sqrt{(1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 3: Calculate the dot product \( \vec{v} \cdot \vec{b} \)
We need to find the dot product of \( \vec{v} \) and \( \vec{b} \):
\[ \vec{v} \cdot \vec{b} = (3\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (\hat{i} - \hat{j} + \hat{k}) \]
Calculating the dot product:
\[ = 3 \cdot 1 + 2 \cdot (-1) + 3 \cdot 1 = 3 - 2 + 3 = 4 \]

### Step 4: Calculate the projection of \( \vec{v} \) onto \( \vec{b} \)
The projection of vector \( \vec{v} \) onto vector \( \vec{b} \) is given by the formula:
\[ \text{proj}_{\vec{b}} \vec{v} = \frac{\vec{v} \cdot \vec{b}}{|\vec{b}|^2} \vec{b} \]
First, we need to calculate \( |\vec{b}|^2 \):
\[ |\vec{b}|^2 = (\sqrt{3})^2 = 3 \]
Now substituting the values:
\[ \text{proj}_{\vec{b}} \vec{v} = \frac{4}{3} \vec{b} \]

### Step 5: Substitute \( \vec{b} \) into the projection formula
Substituting \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \):
\[ \text{proj}_{\vec{b}} \vec{v} = \frac{4}{3} (\hat{i} - \hat{j} + \hat{k}) = \frac{4}{3}\hat{i} - \frac{4}{3}\hat{j} + \frac{4}{3}\hat{k} \]

### Final Answer
Thus, the vector component of the velocity along the line \( \hat{i} - \hat{j} + \hat{k} \) is:
\[ \frac{4}{3}\hat{i} - \frac{4}{3}\hat{j} + \frac{4}{3}\hat{k} \]