Question

If the lines $\dfrac{x-4}{m} = \dfrac{y-3}{2} = \dfrac{z+2}{1}$ and $\dfrac{x-3}{1} = \dfrac{y-4}{1} = \dfrac{z+3}{m}$ are coplanar, then the values of $m$ are

• A. $4,\,1$
• B. $1,\,-4$
• C. $3,\,1$
• D. $5,\,-1$
• E. $3,\,-1$

Hint:

Two lines in 3D are coplanar iff $(\overrightarrow{AB})\cdot(\vec{d_1}\times\vec{d_2})=0$. Set up the $3\times3$ determinant with $\overrightarrow{AB}$, $\vec{d_1}$, $\vec{d_2}$ as rows and equate to zero.

Answer 1

The Correct Option is B

Step 1: Concept:
• Two lines are coplanar if the scalar triple product is zero: \[ \overrightarrow{AB} \cdot (\vec{d_1} \times \vec{d_2}) = 0 \]

Step 2: Identify Vectors:
• Point on Line 1: \(A(4,\,3,\,-2)\), direction: \(\vec{d_1} = (m,\,2,\,1)\)
• Point on Line 2: \(B(3,\,4,\,-3)\), direction: \(\vec{d_2} = (1,\,1,\,m)\)
• Vector joining points: \[ \overrightarrow{AB} = (3-4,\,4-3,\,-3-(-2)) = (-1,\,1,\,-1) \]

Step 3: Apply Determinant Condition:
• Scalar triple product: \[ \begin{vmatrix} -1 & 1 & -1 m & 2 & 1 1 & 1 & m \end{vmatrix} = 0 \]
• Expanding: \[ -1(2m - 1) - 1(m^2 - 1) + (-1)(m - 2) = 0 \]
• Simplify: \[ -2m + 1 - m^2 + 1 - m + 2 = 0 \] \[ -m^2 - 3m + 4 = 0 \]
• Rearranging: \[ m^2 + 3m - 4 = 0 \]

Step 4: Final Answer:
• Factorize: \[ (m + 4)(m - 1) = 0 \]
• \[ m = 1 \quad \text{or} \quad m = -4 \]