Write the direction-cosines of the vector
`hati + 2 hatj + 3 hatk`.

Answer

To find the direction cosines of the vector given by \( \mathbf{v} = \hat{i} + 2\hat{j} + 3\hat{k} \), we will follow these steps:

### Step 1: Identify the components of the vector
The vector can be expressed in terms of its components:
- \( \alpha = 1 \) (coefficient of \( \hat{i} \))
- \( \beta = 2 \) (coefficient of \( \hat{j} \))
- \( \gamma = 3 \) (coefficient of \( \hat{k} \))

### Step 2: Calculate the magnitude of the vector
The magnitude \( |\mathbf{v}| \) of the vector is given by the formula:
\[ |\mathbf{v}| = \sqrt{\alpha^2 + \beta^2 + \gamma^2} \]
Substituting the values:
\[ |\mathbf{v}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Step 3: Calculate the direction cosines
The direction cosines \( l, m, n \) are given by the formulas:
\[ l = \frac{\alpha}{|\mathbf{v}|}, \quad m = \frac{\beta}{|\mathbf{v}|}, \quad n = \frac{\gamma}{|\mathbf{v}|} \]
Substituting the values we calculated:
\[ l = \frac{1}{\sqrt{14}}, \quad m = \frac{2}{\sqrt{14}}, \quad n = \frac{3}{\sqrt{14}} \]

### Step 4: Write down the direction cosines
Thus, the direction cosines of the vector \( \hat{i} + 2\hat{j} + 3\hat{k} \) are:
\[ \left( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right) \]

### Final Answer
The direction cosines of the vector \( \hat{i} + 2\hat{j} + 3\hat{k} \) are:
\[ \left( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right) \]

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