Question

If the plane \[ x - y + z + 4 = 0 \] divides the line joining the points \[ P(2,3,-1) \quad \text{and} \quad Q(1,4,-2) \] in the ratio \( l:m \), then \( l + m \) is:

• A. \( -1 \)
• B. \( 3 \)
• C. \( -1 \)
• D. \( 4 \)

Hint:

For a plane dividing a line segment, apply the section formula in 3D and substitute into the given plane equation.

Answer 1

The Correct Option is B

Step 1: Section formula in 3D Using the section formula, the point dividing \( P \) and \( Q \) in the ratio \( l:m \) is: \[ (x, y, z) = \left(\frac{l x_2 + m x_1}{l+m}, \frac{l y_2 + m y_1}{l+m}, \frac{l z_2 + m z_1}{l+m} \right). \] Solving for the given plane equation, we find: \[ l + m = 3. \]

Answer 2

Given:
- Plane: \[ x - y + z + 4 = 0 \] - Points: \[ P(2, 3, -1), \quad Q(1, 4, -2) \] - The plane divides the line segment \( PQ \) in the ratio \( l : m \).

Step 1: Let the point of division be \( R \), dividing \( PQ \) in ratio \( l : m \) internally.
Coordinates of \( R \) are:
\[ R = \left( \frac{l \cdot x_2 + m \cdot x_1}{l + m}, \frac{l \cdot y_2 + m \cdot y_1}{l + m}, \frac{l \cdot z_2 + m \cdot z_1}{l + m} \right) \] where \( P = (x_1, y_1, z_1) = (2, 3, -1) \), \( Q = (x_2, y_2, z_2) = (1, 4, -2) \).

Step 2: Substitute:
\[ R = \left( \frac{l \cdot 1 + m \cdot 2}{l + m}, \frac{l \cdot 4 + m \cdot 3}{l + m}, \frac{l \cdot (-2) + m \cdot (-1)}{l + m} \right) = \left( \frac{l + 2m}{l + m}, \frac{4l + 3m}{l + m}, \frac{-2l - m}{l + m} \right) \]

Step 3: Since \( R \) lies on the plane \( x - y + z + 4 = 0 \), substitute \( R \) coordinates:
\[ \frac{l + 2m}{l + m} - \frac{4l + 3m}{l + m} + \frac{-2l - m}{l + m} + 4 = 0 \]

Step 4: Multiply both sides by \( l + m \):
\[ (l + 2m) - (4l + 3m) + (-2l - m) + 4(l + m) = 0 \]
Simplify:
\[ l + 2m - 4l - 3m - 2l - m + 4l + 4m = 0 \] \[ (l - 4l - 2l + 4l) + (2m - 3m - m + 4m) = 0 \] \[ (-1l) + (2m) = 0 \] \[ - l + 2 m = 0 \quad \Rightarrow \quad l = 2 m \]

Step 5: We want \( l + m \). Substitute \( l = 2 m \):
\[ l + m = 2m + m = 3 m \]

Since the ratio is \( l : m = 2 : 1 \), the sum is:
\[ l + m = 2 + 1 = 3 \]

Therefore,
\[ \boxed{3} \]