Write the direction ratios of the vector `r=hati-hatj+2hatk` and hence calculate its direction cosines.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the vector
The given vector is:
\[ \mathbf{r} = \hat{i} - \hat{j} + 2\hat{k} \]

### Step 2: Write the direction ratios
The direction ratios of a vector are the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\). For the vector \(\mathbf{r}\):
- The coefficient of \(\hat{i}\) is \(1\),
- The coefficient of \(\hat{j}\) is \(-1\),
- The coefficient of \(\hat{k}\) is \(2\).

Thus, the direction ratios of the vector \(\mathbf{r}\) are:
\[ (1, -1, 2) \]

### Step 3: Calculate the modulus of the vector
The modulus (magnitude) of the vector \(\mathbf{r}\) is calculated using the formula:
\[ |\mathbf{r}| = \sqrt{(1)^2 + (-1)^2 + (2)^2} \]
Calculating this gives:
\[ |\mathbf{r}| = \sqrt{1 + 1 + 4} = \sqrt{6} \]

### Step 4: Calculate the direction cosines
The direction cosines are given by the ratios of the components of the vector to its modulus. Therefore, we calculate:
- Direction cosine along \(x\) (cosine of angle with x-axis):
\[ \cos \alpha = \frac{1}{\sqrt{6}} \]
- Direction cosine along \(y\) (cosine of angle with y-axis):
\[ \cos \beta = \frac{-1}{\sqrt{6}} \]
- Direction cosine along \(z\) (cosine of angle with z-axis):
\[ \cos \gamma = \frac{2}{\sqrt{6}} \]

### Final Result
The direction cosines of the vector \(\mathbf{r}\) are:
\[ \left( \frac{1}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right) \]

### Summary
- Direction Ratios: \( (1, -1, 2) \)
- Direction Cosines: \( \left( \frac{1}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \right) \)

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