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 \(\vec{C} = \vec{A} + \vec{B}\) along the vector \(\vec{B}\). Here are the steps to achieve that:

### Step 1: Find the vector \(\vec{C}\)
Given:
\[ \vec{A} = 2\hat{i} + \hat{j} + \hat{k} \]
\[ \vec{B} = \hat{i} + 2\hat{j} + 2\hat{k} \]
We can find \(\vec{C}\) by adding \(\vec{A}\) and \(\vec{B}\):
\[ \vec{C} = \vec{A} + \vec{B} = (2 + 1)\hat{i} + (1 + 2)\hat{j} + (1 + 2)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k} \]

### Step 2: Calculate the magnitudes of \(\vec{C}\) and \(\vec{B}\)
The magnitude of \(\vec{C}\) is calculated as follows:
\[ |\vec{C}| = \sqrt{(3)^2 + (3)^2 + (3)^2} = \sqrt{27} = 3\sqrt{3} \]

The magnitude of \(\vec{B}\) is calculated as follows:
\[ |\vec{B}| = \sqrt{(1)^2 + (2)^2 + (2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Step 3: Calculate the dot product \(\vec{C} \cdot \vec{B}\)
Now, we find the dot product \(\vec{C} \cdot \vec{B}\):
\[ \vec{C} \cdot \vec{B} = (3\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) \]
Calculating the dot product:
\[ = 3 \cdot 1 + 3 \cdot 2 + 3 \cdot 2 = 3 + 6 + 6 = 15 \]

### Step 4: Calculate \(\cos \theta\)
Using the formula for the cosine of the angle \(\theta\) between \(\vec{C}\) and \(\vec{B}\):
\[ \cos \theta = \frac{\vec{C} \cdot \vec{B}}{|\vec{C}| |\vec{B}|} = \frac{15}{(3\sqrt{3})(3)} = \frac{15}{9\sqrt{3}} = \frac{5}{3\sqrt{3}} \]

### Step 5: Find the component of \(\vec{C}\) along \(\vec{B}\)
The component of \(\vec{C}\) along \(\vec{B}\) is given by:
\[ \text{Component of } \vec{C} \text{ along } \vec{B} = |\vec{C}| \cos \theta \]
Substituting the values:
\[ = (3\sqrt{3}) \left(\frac{5}{3\sqrt{3}}\right) = 5 \]

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

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