Find the unit vector in the direction of the vector :
`vec(a)=2hat(i)+3hat(j)+hat(k)`

Answer

To find the unit vector in the direction of the vector \(\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}\), we can follow these steps:

### Step 1: Identify the vector
The given vector is:
\[ \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k} \]

### Step 2: Calculate the magnitude of the vector
The magnitude (or modulus) of the vector \(\vec{a}\) is calculated using the formula:
\[ |\vec{a}| = \sqrt{(a_x^2 + a_y^2 + a_z^2)} \]
where \(a_x\), \(a_y\), and \(a_z\) are the components of the vector.

For \(\vec{a}\):
- \(a_x = 2\)
- \(a_y = 3\)
- \(a_z = 1\)

Now, substituting the values:
\[ |\vec{a}| = \sqrt{(2^2 + 3^2 + 1^2)} = \sqrt{(4 + 9 + 1)} = \sqrt{14} \]

### Step 3: Calculate the unit vector
The unit vector \(\hat{a}\) in the direction of \(\vec{a}\) is given by:
\[ \hat{a} = \frac{\vec{a}}{|\vec{a}|} \]

Substituting the values we have:
\[ \hat{a} = \frac{2\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{14}} \]

### Step 4: Express the unit vector in component form
This can be expressed as:
\[ \hat{a} = \frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} + \frac{1}{\sqrt{14}}\hat{k} \]

### Final Answer
Thus, the unit vector in the direction of the vector \(\vec{a}\) is:
\[ \hat{a} = \frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} + \frac{1}{\sqrt{14}}\hat{k} \]

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