Let the line L pass through (1, 1, 1) and intersect the lines and . Then which of the following points lies on the line L?

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Solution

Q.
Let the line L pass through (1, 1, 1) and intersect the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}$ . Then which of the following points lies on the line L [2025]

1 (10, –29, –50)

2 (4, 22, 7)

3 (7, 15, 13)

4 (5, 4, 3)

Ans.
(3)

Equation of line L passes through (1, 1, 1) is $L : \frac{x - 1}{a} = \frac{y - 1}{b} = \frac{z - 1}{c}$ , where a, b, c are direction ratios.

Let $L_{1} : \frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4} = λ (say)$

Any point on $L_{1}$ be $A (2 λ + 1, 3 λ - 1, 4 λ + 1)$

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

Any point on $L_{2}$ be $B (μ + 3, 2 μ + 4, μ)$

Direction ratio of L be : $< 2 λ, 3 λ - 2, 4 λ > or < μ + 2, 2 μ + 3, μ - 1 >$

Now, $\frac{2 λ}{μ + 2} = \frac{3 λ - 2}{2 μ + 3} = \frac{4 λ}{μ - 1} \Rightarrow λ = \frac{- 6}{5} and μ = - 5$

$∴ < a, b, c > \equiv < - 3, - 7, - 6 > or < 3, 7, 6 >$

$∴ L : \frac{x - 1}{3} = \frac{y - 1}{7} = \frac{z - 1}{6}$

Hence, (7, 15, 13) lies on the line.

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