Question

If the lines \(\frac{x-1}{2} = \frac{y+1}{k} = \frac{z}{2}\) and \(\frac{x+1}{5} = \frac{y+1}{2} = \frac{z}{k}\) are coplanar, then the equation of the plane containing these lines are

• A. \(x \pm y + z = 0\)
• B. \(y \pm z + 1 = 0\)
• C. \(2x \pm y = 0\)
• D. \(x \pm z + 1 = 0\)

Hint:

For coplanar-line questions in 3D, first find the allowed parameter values from the scalar triple product, then determine the plane.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Two lines are coplanar if the scalar triple product of the difference between points on the lines and their direction vectors is zero.
Step 2: Key Formula or Approach:
1. Coplanarity condition: \(\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1
l_1 & m_1 & n_1
l_2 & m_2 & n_2 \end{vmatrix} = 0\).
2. Plane containing lines has a normal \(\bar{n} = \bar{v_1} \times \bar{v_2}\).
Step 3: Detailed Explanation:
Points on lines are \(P_1(1, -1, 0)\) and \(P_2(-1, -1, 0)\). Difference vector \(= (-2, 0, 0)\).
Direction ratios are \((2, k, 2)\) and \((5, 2, k)\).
Using the determinant:
\[ \begin{vmatrix} -2 & 0 & 0
2 & k & 2
5 & 2 & k \end{vmatrix} = -2(k^2 - 4) = 0 \implies k = \pm 2 \] Case I: \(k = 2\). Direction vectors are \((2, 2, 2)\) and \((5, 2, 2)\).
Normal \(\bar{n} = (2, 2, 2) \times (5, 2, 2) = (0, 6, -6)\). Simplified DR \((0, 1, -1)\).
Plane: \(0(x-1) + 1(y+1) - 1(z) = 0 \implies y - z + 1 = 0\).
Case II: \(k = -2\). Direction vectors are \((2, -2, 2)\) and \((5, 2, -2)\).
Normal \(\bar{n} = (2, -2, 2) \times (5, 2, -2) = (0, 14, 14)\). Simplified DR \((0, 1, 1)\).
Plane: \(0(x-1) + 1(y+1) + 1(z) = 0 \implies y + z + 1 = 0\).
Combined: \(y \pm z + 1 = 0\).
Step 4: Final Answer:
The plane equation is \(y \pm z + 1 = 0\).