Let and be two lines. Let be a line passing through the point and be perpendicular to both and . If intersects , then equals : [2025]

Question

Let $L_{1} : \frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z - 1}{2}$ and $L_{2} : \frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z}{1}$ be two lines.

Let $L_{3}$ be a line passing through the point $(α, β, γ)$ and be perpendicular to both $L_{1}$ and $L_{2}$ . If $L_{3}$ intersects $L_{1}$ , then $| 5 α - 11 β - 8 γ |$ equals : [2025]

Smart Achievers · Accepted Answer

(2)

Let D.r.s. of $L_{3}$ be a, b, c

Now, $L_{3}$ is $⊥^{r}$ to $L_{1}$ and $L_{2}$ = a – b + 2c = 0 ... (i)

and –a + 2b + c = 0 ... (ii)

Solving (i) and (ii), we get $\frac{a}{- 5} = \frac{b}{- 3} = \frac{c}{1}$

Equation of line $L_{3}$ is $\frac{x - α}{- 5} = \frac{y - β}{- 3} = \frac{z - γ}{1} = k$

Any point on $L_{3}$ is $(- 5 k + α, - 3 k + β, k + γ)$

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

Any point on $L_{1}$ is $(λ + 1, - λ + 2, 2 λ + 1)$

Now $L_{1}$ and $L_{3}$ intersects.

$∴ - 5 k + α = λ + 1, - 3 k + β = - λ + 2, k + γ = 2 λ + 1$

$\Rightarrow α = 5 k + λ + 1, β = 3 k - λ + 2, γ = - k + 2 λ + 1$

$∴ | 5 α - 11 β - 8 γ |$

$= | 25 k + 5 λ + 5 - 33 k + 11 λ - 22 + 8 k - 16 λ - 8 | = | - 25 | = 25$ .

Solution

1 20

2 25

3 16

4 18

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