Let the position vectors of the points P, A and B be `r,i+j+k` and `-i+k`. If PA is perpendicular to PB but is not perpendicular to `r-(j+2k)`, then r is :

Options

A. `i+2k`
B. `i+2j`
C. `j-2k`
D. `j+2k`

Answer

To solve the problem, we need to analyze the given position vectors and the conditions provided. Let's break it down step by step.

### Given:
- Position vector of point P: \( \vec{r} = \hat{i} + \hat{j} + \hat{k} \)
- Position vector of point A: \( \vec{a} = -\hat{i} + \hat{k} \)
- Position vector of point B: \( \vec{b} = -\hat{i} + \hat{k} \)

### Step 1: Find the vectors PA and PB
The vector \( \vec{PA} \) is given by:
\[ \vec{PA} = \vec{a} - \vec{r} = (-\hat{i} + \hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = -2\hat{i} - \hat{j} \]

The vector \( \vec{PB} \) is given by:
\[ \vec{PB} = \vec{b} - \vec{r} = (-\hat{i} + \hat{k}) - (\hat{i} + \hat{j} + \hat{k}) = -2\hat{i} - \hat{j} \]

### Step 2: Check the condition for perpendicularity
For \( \vec{PA} \) to be perpendicular to \( \vec{PB} \), their dot product must equal zero:
\[ \vec{PA} \cdot \vec{PB} = (-2\hat{i} - \hat{j}) \cdot (-2\hat{i} - \hat{j}) = 4 + 1 = 5 \neq 0 \]
This indicates that \( \vec{PA} \) is not perpendicular to \( \vec{PB} \).

### Step 3: Analyze the second condition
We need to check the condition that \( \vec{PA} \) is not perpendicular to \( \vec{r} - (\hat{j} + 2\hat{k}) \):
\[ \vec{r} - (\hat{j} + 2\hat{k}) = (\hat{i} + \hat{j} + \hat{k}) - (\hat{j} + 2\hat{k}) = \hat{i} - \hat{k} \]

Now, we compute the dot product:
\[ \vec{PA} \cdot (\hat{i} - \hat{k}) = (-2\hat{i} - \hat{j}) \cdot (\hat{i} - \hat{k}) = -2 + 0 = -2 \neq 0 \]
This confirms that \( \vec{PA} \) is not perpendicular to \( \vec{r} - (\hat{j} + 2\hat{k}) \).

### Step 4: Conclusion
Since both conditions are satisfied, we conclude that the position vector \( \vec{r} \) is:
\[ \vec{r} = \hat{i} + \hat{j} + \hat{k} \]