If points `hati+hatj, hati -hatj and p hati +qhatj+rhatk` are collinear, then

Options

A. `p=1`
B. `r=0`
C. `q in R`
D. `q ne 1`

Explanation

To determine the values of \( p \), \( q \), and \( r \) such that the points \( \hat{i} + \hat{j} \), \( \hat{i} - \hat{j} \), and \( p\hat{i} + q\hat{j} + r\hat{k} \) are collinear, we can follow these steps:

### Step 1: Define the Points
Let:
- Point A = \( \hat{i} + \hat{j} \)
- Point B = \( \hat{i} - \hat{j} \)
- Point C = \( p\hat{i} + q\hat{j} + r\hat{k} \)

### Step 2: Find the Vectors AB and BC
To check for collinearity, we need to find the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \).

1. **Vector \( \overrightarrow{AB} \)**:
\[ \overrightarrow{AB} = B - A = (\hat{i} - \hat{j}) - (\hat{i} + \hat{j}) = \hat{i} - \hat{j} - \hat{i} - \hat{j} = -2\hat{j} \]

2. **Vector \( \overrightarrow{BC} \)**:
\[ \overrightarrow{BC} = C - B = (p\hat{i} + q\hat{j} + r\hat{k}) - (\hat{i} - \hat{j}) = (p - 1)\hat{i} + (q + 1)\hat{j} + r\hat{k} \]

### Step 3: Set Up the Collinearity Condition
For points A, B, and C to be collinear, vectors \( \overrightarrow{AB} \) and \( \overrightarrow{BC} \) must be parallel. This implies that the ratios of their components must be equal.

From the vectors:
- \( \overrightarrow{AB} = -2\hat{j} \) implies the \( \hat{i} \) component is 0 and the \( \hat{j} \) component is -2.
- For \( \overrightarrow{BC} \) to be parallel to \( \overrightarrow{AB} \):
\[ \frac{p - 1}{0} = \frac{q + 1}{-2} = \frac{r}{0} \]

### Step 4: Solve for \( p \), \( q \), and \( r \)
1. From \( \frac{p - 1}{0} \): This implies \( p - 1 = 0 \) (since the \( \hat{i} \) component must be 0). Thus:
\[ p = 1 \]

2. From \( \frac{q + 1}{-2} = 1 \):
\[ q + 1 = -2 \implies q = -3 \]

3. From \( \frac{r}{0} \): This implies \( r = 0 \) (since the \( \hat{k} \) component must be 0).

### Final Values
Thus, we have:
- \( p = 1 \)
- \( q = -3 \)
- \( r = 0 \)

### Summary
The values of \( p \), \( q \), and \( r \) are:
\[ p = 1, \quad q = -3, \quad r = 0 \]