Question

Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $

• A. \( 8 \)
• B. \( 16 \)
• C. \( 12 \)
• D. \( 6 \)

Hint:

To check whether two vectors are collinear, equating the ratios of their direction ratios is a powerful approach. Also, using parametric representation for lines makes it easier to substitute and calculate points of intersection.

Answer 1

The Correct Option is A

 
The line passing through point \( P(4,1,0) \) intersects line \( L_1 \) and \( L_2 \). 
Let us assume the point \( A \) on \( L_1 \) is given by parameter \( p \): \[ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = p \Rightarrow A(2p + 1, 3p + 2, 4p + 3) \] Let point \( B \) on \( L_2 \) be parameterized using \( q \): \[ L_2: x - 6 = y = -z + 4 = q \Rightarrow B(q + 6, q, 4 - q) \] Now find the direction ratios (D.R.) of \( \overrightarrow{PA} \) and \( \overrightarrow{PB} \): \[ \text{D.R. of } \overrightarrow{PA} = (2p - 3, 3p + 1, 4p + 3) \text{D.R. of } \overrightarrow{PB} = (q + 2, q - 1, 4 - q) \] Since \( \overrightarrow{PA} \) and \( \overrightarrow{PB} \) are collinear, their components must be proportional: \[ \frac{2p - 3}{q + 2} = \frac{3p + 1}{q - 1} = \frac{4p + 3}{4 - q} \] Equating pairwise and solving: \[ 2pq - 2p - 3q + 3 = 3pq + 6p + q + 2 \Rightarrow pq + rp + 4q - 1 = 0 \quad \text{(1)} 7pq - 16p + 4q - 7 = 0 \quad \text{(2)} \] Solving these equations, we find: \[ pq = -3, \quad p = -1, \quad q = 3 \Rightarrow A(-1, -1, -1), \quad B(9, 3, 1) \] Now compute the determinant: \[ \begin{vmatrix} 1 & 0 & 1 -1 & -1 & -1 \\ 9 & 3 & 1 \end{vmatrix} = 1 \cdot \begin{vmatrix}-1 & -1 \\ 3 & 1 \end{vmatrix} - 0 + 1 \cdot \begin{vmatrix}-1 & -1 \\ 9 & 3 \end{vmatrix} = 1(-1 + 3) + 1( -3 + 9 ) = 2 + 6 = 8 \]