If `a = i+2j+2k`, and `b=3i+6j+2k`, then the vector in the direction of a and having magnitude of b is.......

Answer

To find the vector in the direction of vector \( \mathbf{a} \) with the magnitude of vector \( \mathbf{b} \), we can follow these steps:

### Step 1: Identify the vectors
Given:
\[ \mathbf{a} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \]
\[ \mathbf{b} = 3\mathbf{i} + 6\mathbf{j} + 2\mathbf{k} \]

### Step 2: Find the magnitude of vector \( \mathbf{b} \)
The magnitude of a vector \( \mathbf{b} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \) is given by:
\[ |\mathbf{b}| = \sqrt{b_1^2 + b_2^2 + b_3^2} \]
For \( \mathbf{b} \):
\[ |\mathbf{b}| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 \]

### Step 3: Find the unit vector in the direction of \( \mathbf{a} \)
The unit vector \( \hat{\mathbf{a}} \) in the direction of \( \mathbf{a} \) is given by:
\[ \hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} \]
First, we need to find the magnitude of \( \mathbf{a} \):
\[ |\mathbf{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]
Now, we can find the unit vector:
\[ \hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}}{3} = \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k} \]

### Step 4: Scale the unit vector by the magnitude of \( \mathbf{b} \)
To find the vector in the direction of \( \mathbf{a} \) with the magnitude of \( \mathbf{b} \), we multiply the unit vector \( \hat{\mathbf{a}} \) by the magnitude of \( \mathbf{b} \):
\[ \text{Desired vector} = |\mathbf{b}| \cdot \hat{\mathbf{a}} = 7 \left( \frac{1}{3}\mathbf{i} + \frac{2}{3}\mathbf{j} + \frac{2}{3}\mathbf{k} \right) \]
Calculating this gives:
\[ \text{Desired vector} = \frac{7}{3}\mathbf{i} + \frac{14}{3}\mathbf{j} + \frac{14}{3}\mathbf{k} \]

### Final Answer
The vector in the direction of \( \mathbf{a} \) and having the magnitude of \( \mathbf{b} \) is:
\[ \frac{7}{3}\mathbf{i} + \frac{14}{3}\mathbf{j} + \frac{14}{3}\mathbf{k} \]

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