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

Concept:
Two lines are coplanar if the scalar triple product of:
• a vector joining one point from each line,
• direction vector of the first line,
• direction vector of the second line e} is zero. ip

Step 1: Write points and direction vectors.
From the first line: \[ P_1=(1,-1,0),\qquad \vec{d}_1=(2,k,2) \] From the second line: \[ P_2=(-1,-1,0),\qquad \vec{d}_2=(5,2,k) \] A joining vector is: \[ \overrightarrow{P_1P_2}=(-2,0,0) \] ip

Step 2: Apply the coplanarity condition.
\[ \begin{vmatrix} -2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k \end{vmatrix} =0 \] Expanding: \[ -2(k^2-4)=0 \] \[ k^2=4 \] \[ k=\pm 2 \] ip

Step 3: Find the plane for each value of \(k\).
If \(k=2\), the plane turns out to be: \[ y-z+1=0 \] If \(k=-2\), the plane turns out to be: \[ y+z+1=0 \] So the plane containing the lines is: \[ y\pm z+1=0 \] ip Hence, the correct answer is:
\[ \boxed{(B)\ y\pm z+1=0} \]