If `vec(A)=2hat(i)+hat(j)+hat(k)` and `vec(B)=hat(i)+2hat(j)+2hat(k)`, find the magnitude of compinent of `(vec(A)+vec(B))` along `vec(B)`

Options

A. `4` unit
B. `5` unit
C. `6` unit
D. `7` unit

Explanation

To solve the problem, we need to find the magnitude of the component of the vector \(\vec{A} + \vec{B}\) along the vector \(\vec{B}\).

### Step-by-Step Solution:

1. **Define the vectors**:
\[ \vec{A} = 2\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{B} = \hat{i} + 2\hat{j} + 2\hat{k} \]

2. **Calculate \(\vec{A} + \vec{B}\)**:
\[ \vec{A} + \vec{B} = (2\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} + 2\hat{k}) \]
\[ = (2 + 1)\hat{i} + (1 + 2)\hat{j} + (1 + 2)\hat{k} \]
\[ = 3\hat{i} + 3\hat{j} + 3\hat{k} \]

3. **Find the magnitude of \(\vec{A} + \vec{B}\)**:
\[ |\vec{A} + \vec{B}| = \sqrt{(3)^2 + (3)^2 + (3)^2} = \sqrt{27} = 3\sqrt{3} \]

4. **Calculate the magnitude of \(\vec{B}\)**:
\[ |\vec{B}| = \sqrt{(1)^2 + (2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

5. **Calculate the dot product \((\vec{A} + \vec{B}) \cdot \vec{B}\)**:
\[ (\vec{A} + \vec{B}) \cdot \vec{B} = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) \]
\[ = 3 \cdot 1 + 3 \cdot 2 + 3 \cdot 2 = 3 + 6 + 6 = 15 \]

6. **Find the component of \(\vec{A} + \vec{B}\) along \(\vec{B}\)**:
The component of \(\vec{A} + \vec{B}\) along \(\vec{B}\) is given by:
\[ \text{Component} = \frac{(\vec{A} + \vec{B}) \cdot \vec{B}}{|\vec{B}|} \]
\[ = \frac{15}{3} = 5 \]

7. **Magnitude of the component**:
Since we are asked for the magnitude of the component, it is simply:
\[ \text{Magnitude of the component} = 5 \]

### Final Answer:
The magnitude of the component of \((\vec{A} + \vec{B})\) along \(\vec{B}\) is \(5\).