Question

The equation of the plane containing the lines \(\frac{x-1}{2} = \frac{y+1}{λ} = \frac{z}{2}\) and \(\frac{x+1}{5} = \frac{y+1}{2} = \frac{z}{λ}\) is

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

Hint:

Coplanarity gives lambda = plus or minus 2. Then cross the directions.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Two lines lie in one plane (coplanar) if the vector joining one point of each, and the two direction vectors, have zero scalar triple product.

Step 2: Data:
Line 1: point \((1,-1,0)\), direction \((2,\lambda,2)\). Line 2: point \((-1,-1,0)\), direction \((5,2,\lambda)\). Joining vector: \((-2,0,0)\).

Step 3: Coplanarity:
\[ \begin{vmatrix}-2&0&0\\2&\lambda&2\\5&2&\lambda\end{vmatrix}=-2(\lambda^2-4)=0\ \Rightarrow\ \lambda=\pm2 \]

Step 4: Case lambda = 2:
\(\vec d_1=(2,2,2)\) and \(\vec d_2=(5,2,2)\). Normal \(=\vec d_1\times\vec d_2=(0,6,-6)\), so the plane is \(y-z=\text{const}\). At \((1,-1,0)\) the constant is \(-1\). So \(y-z+1=0\).

Step 5: Case lambda = -2:
\(\vec d_1=(2,-2,2)\) and \(\vec d_2=(5,2,-2)\). Normal \(=(0,14,14)\), so \(y+z=\text{const}=-1\). So \(y+z+1=0\).

Step 6: Choose:
Both cases combine as \(y\pm z+1=0\), option (B).

Final Answer:
The plane is y plus or minus z plus 1 = 0. \[ \boxed{y\pm z+1=0} \]