Let the line L intersect the lines x – 2 = – y = z – 1, 2(x + 1) = 2(y –1) = z + 1 and be parallel to the line . Then which of the following points lis on L? [2024]

Question

Let the line L intersect the lines x – 2 = – y = z – 1, 2(x + 1) = 2(y –1) = z + 1 and be parallel to the line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}$ . Then which of the following points lis on L? [2024]

Smart Achievers · Accepted Answer

(2)

$L_{1} : \frac{x - 2}{1} = \frac{y}{- 1} = \frac{z - 1}{1} = λ$

$L_{2} : \frac{x + 1}{(1 / 2)} = \frac{y - 1}{(1 / 2)} = \frac{z + 1}{1} = μ$

Any point on $L_{1}$ and $L_{2}$ will be of the form $(λ + 2, - λ, λ + 1)$ and $(\frac{μ}{2} - 1, \frac{μ}{2} + 1, μ - 1)$ respectively.

Now direction ratios of line L are given by

$< λ - \frac{μ}{2} + 3, - λ - \frac{μ}{2} - 1, λ - μ + 2 >$

Also L is parallel to line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}$

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

$\begin{matrix} \Rightarrow λ - \frac{μ}{2} + 3 = - 3 λ - \frac{3 μ}{2} - 3 . . . (i) \\ 2 (λ - \frac{μ}{2} + 3) = 3 (λ - μ + 2) . . . (ii) \end{matrix}} λ = \frac{- 4}{3}, μ = \frac{- 2}{3}$

$∴$ Points will be $(\frac{2}{3}, \frac{4}{3}, \frac{- 1}{3})$ and $(\frac{- 4}{3}, \frac{2}{3}, \frac{- 5}{3})$

$∴$ L is given by $\frac{x - \frac{2}{3}}{3} = \frac{y - \frac{4}{3}}{1} = \frac{z + \frac{1}{3}}{2}$

Point $(\frac{- 1}{3}, 1, - 1)$ will satisfy this line L.

Solution

Q.
Let the line L intersect the lines x – 2 = – y = z – 1, 2(x + 1) = 2(y –1) = z + 1 and be parallel to the line $\frac{x - 2}{3} = \frac{y - 1}{1} = \frac{z - 2}{2}$ . Then which of the following points lis on L [2024]

1 $(- \frac{1}{3}, - 1, - 1)$

2 $(- \frac{1}{3}, 1, - 1)$

3 $(- \frac{1}{3}, - 1, 1)$

4 $(- \frac{1}{3}, 1, 1)$

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Solution

Q. Let the line L intersect the lines x – 2 = – y = z – 1, 2(x + 1) = 2(y –1) = z + 1 and be parallel to the line x–23=y–11=z–22. Then which of the following points lis on L [2024]

1 (–13,–1,–1)

2 (–13,1,–1)

3 (–13,–1,1)