Let a line passing through the point (4, 1, 0) intersect the line L1: (x-1)/2 = (y-2)/3 = (z-3)/4 at the point A(alpha, beta, gamma) and the line L2: x-6 = y = -z+4 at the point B(a, b, c). Then | 1 0 1 ; alpha beta gamma ; a b c | is equal to

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q15 – JEE-MAIN 2025

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Home Maths 3D GeometryQuestion #22

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 (α, β, γ)

and the line

L_{2} : x - 6 = y = - z + 4

at the point

B (a, b, c)

. Then

| \begin{matrix} 1 & 0 & 1 \\ α & β & γ \\ a & b & c \end{matrix} |

is equal to

Clue 1 of 3: Collinearity Condition

The problem states that a single line passes through point $P (4, 1, 0)$ and intersects $L_{1}$ at $A$ and $L_{2}$ at $B$ . This implies that points $P, A, B$ are collinear.

🧠 Full Solution Path

Step 1: Determine Coordinates of A and B using Collinearity✦ Active

Let the given point be $P (4, 1, 0)$ . Any point $A$ on line $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$ can be written as $A (2 λ + 1, 3 λ + 2, 4 λ + 3)$ . Any point $B$ on line $L_{2} : x - 6 = y = - z + 4 = μ$ can be written as $B (μ + 6, μ, - μ + 4)$ . Since $P, A, B$ are collinear, the vectors $\vec{P A}$ and $\vec{P B}$ must be parallel. The components of $\vec{P A}$ are $(2 λ - 3, 3 λ + 1, 4 λ + 3)$ and $\vec{P B}$ are $(μ + 2, μ - 1, - μ + 4)$ . Equating the ratios of corresponding components:

\frac{2 λ - 3}{μ + 2} = \frac{3 λ + 1}{μ - 1} = \frac{4 λ + 3}{- μ + 4}

This leads to a system of equations:

{\begin{cases} (2 λ - 3) (μ - 1) = (3 λ + 1) (μ + 2) ⟹ λ μ + 8 λ + 4 μ - 1 = 0 (1) \\ (3 λ + 1) (- μ + 4) = (4 λ + 3) (μ - 1) ⟹ 7 λ μ - 16 λ + 4 μ - 7 = 0 (2) \end{cases}

Solving these equations (e.g., $7 \times (1) - (2)$ ), we get $72 λ + 24 μ = 0 ⟹ μ = - 3 λ$ . Substituting this into (1):

λ (- 3 λ) + 8 λ + 4 (- 3 λ) - 1 = 0 ⟹ - 3 λ^{2} - 4 λ - 1 = 0 ⟹ 3 λ^{2} + 4 λ + 1 = 0

Factoring gives $(3 λ + 1) (λ + 1) = 0$ , so $λ = - 1$ or $λ = - 1 / 3$ . We check for collinearity for both cases. For $λ = - 1 / 3$ , $μ = 1$ . $\vec{P A} = (- 11 / 3, 0, 5 / 3)$ and $\vec{P B} = (3, 0, 3)$ . These vectors are not parallel (e.g., $- 11 / 3 \neq 3 k$ and $5 / 3 \neq 3 k$ for the same $k$ ), so $P, A, B$ are not collinear. Thus, $λ = - 1$ is the only valid solution. For $λ = - 1$ , $μ = - 3 (- 1) = 3$ .

Step 2: Calculate Coordinates of A and B○ Expand

Using $λ = - 1$ :

A (α, β, γ) = (2 (- 1) + 1, 3 (- 1) + 2, 4 (- 1) + 3) = (- 1, - 1, - 1)

Using $μ = 3$ :

B (a, b, c) = (3 + 6, 3, - 3 + 4) = (9, 3, 1)

💡 Teacher's Secret Hint

Ensure to verify the collinearity condition for all possible solutions of lambda and mu.

Step 3: Evaluate the Determinant○ Expand

Substitute the coordinates of $A$ and $B$ into the determinant:

| \begin{matrix} 1 & 0 & 1 \\ α & β & γ \\ a & b & c \end{matrix} | = | \begin{matrix} 1 & 0 & 1 \\ - 1 & - 1 & - 1 \\ 9 & 3 & 1 \end{matrix} |

Expand the determinant along the first row:

D = 1 ((- 1) (1) - (- 1) (3)) - 0 + 1 ((- 1) (3) - (- 1) (9))

Calculate the value:

D = 1 (- 1 + 3) + 1 (- 3 + 9) = 1 (2) + 1 (6) = 2 + 6 = 8

💡 Teacher's Secret Hint

Double-check determinant calculations to avoid arithmetic errors.

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Maths Question 15 – JEE-MAIN 2025

Clue 1 of 3: Collinearity Condition

🧠 Full Solution Path