Let a line passing through the point (4, 1, 0) intersect the line at the point and the line at the point B(a, b, c). Then is equal to [2025]

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 $A (α, β, γ)$ 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 [2025]

Smart Achievers · Accepted Answer

(3)

$L_{1} = \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = t (say)$

$L_{2} = \frac{x - 6}{1} = \frac{y}{1} = \frac{z - 4}{- 1} = μ (say)$

A(2t + 1, 3t + 2, 4t + 3) is any point on $L_{1}$ .

And $B (μ + 6, μ, 4 - μ)$ is any point on $L_{2}$ .

$∴$ D.R. of PA = 2t – 3, 3t + 1, 4t + 3 and D.R. of $P B = μ + 2, μ - 1, 4 - μ$ , where the coordinate of P is (4, 1, 0).

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

[Direction ratios of PA and PB are proportional]

$2 t μ - 2 t - 3 μ + 3 = 3 t μ + 6 t + μ + 2$

$\Rightarrow t μ + 8 t + 4 μ - 1 = 0$ ... (i)

Also, $12 t - 3 μ t + 4 - μ = 4 μ t + 3 μ - 4 t - 3$

$\Rightarrow 7 μ t - 16 t + 4 μ - 7 = 0$ ... (ii)

And $8 t - 2 μ t - 12 + 3 μ = 4 μ t + 8 t + 3 μ + 6$

$\Rightarrow 6 μ t = - 18 \Rightarrow μ t = - 3$

From equation (i), we get

$8 t + 4 μ = 4 \Rightarrow 2 t + μ = 1$ ... (iii)

From equation (ii), we get

$- 21 - 16 t + 4 μ - 7 = 0$

$16 t - 4 μ = - 28 \Rightarrow 4 t - μ = - 7$ ... (iv)

From equation (iii) and (iv), we get

$∴ t = - 1, μ = 3$

$∴ A (- 1, - 1, - 1) and B (9, 3, 1)$

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

1(–1 + 3) – 0 + 1(–3 + 9) = 8.

Solution

1 12

2 6

3 8

4 16

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Solution

Q. Let a line passing through the point (4, 1, 0) intersect the line L1:x–12=y–23=z–34 at the point A(α,β,γ) and the line L2:x–6=y=–z+4 at the point B(a, b, c). Then |101αβγabc| is equal to [2025]

1 12

2 6

3 8

4 16

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