From a point A with position vector `p(i+j+k)`, AB and AC are drawn perpendicular to the lines `r=k+lambda(i+j)" and "r=-k+mu(i-j)` respectively. A value of p is equal to

Options

A. -1
B. `sqrt(2)`
C. 2
D. -2

Answer

To solve the problem, we need to find the value of \( p \) given the position vector of point \( A \) and the equations of two lines. Let's go through the solution step by step.

### Step 1: Identify the position vector of point A
The position vector of point \( A \) is given as:
\[ \vec{A} = p(\hat{i} + \hat{j} + \hat{k}) = p\hat{i} + p\hat{j} + p\hat{k} \]

### Step 2: Write the equations of the lines
The equations of the two lines are:
1. Line \( L_1: \vec{r} = \hat{k} + \lambda(\hat{i} + \hat{j}) \)
2. Line \( L_2: \vec{r} = -\hat{k} + \mu(\hat{i} - \hat{j}) \)

### Step 3: Find the coordinates of point B on line \( L_1 \)
Since point \( B \) lies on line \( L_1 \), we can express its coordinates as:
\[ \vec{B} = \lambda \hat{i} + \lambda \hat{j} + \hat{k} \]

### Step 4: Find the vector \( \vec{AB} \)
The vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = \vec{B} - \vec{A} = (\lambda - p)\hat{i} + (\lambda - p)\hat{j} + (1 - p)\hat{k} \]

### Step 5: Determine the direction vector of line \( L_1 \)
The direction vector of line \( L_1 \) is:
\[ \vec{d_1} = \hat{i} + \hat{j} \]

### Step 6: Set up the perpendicularity condition
Since \( \vec{AB} \) is perpendicular to \( \vec{d_1} \), we have:
\[ \vec{AB} \cdot \vec{d_1} = 0 \]
This gives us:
\[ (\lambda - p) \cdot 1 + (\lambda - p) \cdot 1 + (1 - p) \cdot 0 = 0 \]
Simplifying this, we get:
\[ 2(\lambda - p) = 0 \implies \lambda - p = 0 \implies \lambda = p \]

### Step 7: Find the coordinates of point C on line \( L_2 \)
Point \( C \) lies on line \( L_2 \), so we can express its coordinates as:
\[ \vec{C} = \mu \hat{i} - \mu \hat{j} - \hat{k} \]

### Step 8: Find the vector \( \vec{AC} \)
The vector \( \vec{AC} \) can be calculated as:
\[ \vec{AC} = \vec{C} - \vec{A} = (\mu - p)\hat{i} + (-\mu - p)\hat{j} + (-1 - p)\hat{k} \]

### Step 9: Determine the direction vector of line \( L_2 \)
The direction vector of line \( L_2 \) is:
\[ \vec{d_2} = \hat{i} - \hat{j} \]

### Step 10: Set up the perpendicularity condition for \( \vec{AC} \)
Since \( \vec{AC} \) is perpendicular to \( \vec{d_2} \), we have:
\[ \vec{AC} \cdot \vec{d_2} = 0 \]
This gives us:
\[ (\mu - p) \cdot 1 + (-\mu - p)(-1) + (-1 - p) \cdot 0 = 0 \]
Simplifying this, we get:
\[ (\mu - p) + (\mu + p) = 0 \implies 2\mu = 0 \implies \mu = 0 \]

### Step 11: Conclusion
Since we found that \( \lambda = p \) and \( \mu = 0 \), we can conclude that the value of \( p \) is:
\[ \boxed{0} \]