Let P be the foot of the perpendicular from the point (1, 2, 2) on the line . Let the line , intersect the line L at Q. Then is equal to : [2025]

Question

Let P be the foot of the perpendicular from the point (1, 2, 2) on the line $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2}$ . Let the line $\vec{r} = (- \hat{i} + \hat{j} - 2 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}), λ \in R$ , intersect the line L at Q. Then $2 {(P Q)}^{2}$ is equal to : [2025]

Smart Achievers · Accepted Answer

(3)

We have, $L : \frac{x - 1}{1} = \frac{y + 1}{- 1} = \frac{z - 2}{2} = μ$

$\Rightarrow Point is P (μ + 1, - μ - 1, 2 μ + 2)$

$∴ D . r' s of A P = (μ, - μ - 3, 2 μ)$

Now, $μ (1) + (- μ - 3) (- 1) + 2 (2 μ) = 0$

$\Rightarrow μ + μ + 3 + 4 μ = 0$

$\Rightarrow 6 μ + 3 = 0 \Rightarrow μ = \frac{- 1}{2}$

$∴ P \equiv (\frac{1}{2}, \frac{- 1}{2}, 1)$

Now, line L intersect the other line at Q, i.e.,

$(- 1 + λ, 1 - λ, - 2 + λ)$

$∴ μ + 1 = - 1 + λ, - μ - 1 = 1 - λ and 2 μ + 2 = - 2 + λ$

$\Rightarrow μ = λ - 2, μ = λ - 2 and 2 μ + 2 = - 2 + λ$

$\Rightarrow 2 (λ - 2) + 2 = - 2 + λ$

$\Rightarrow 2 λ - 4 + 2 = - 2 + λ \Rightarrow λ = 0 and μ = - 2$

$∴ Q \equiv (- 1, 1, - 2)$

So, ${(P Q)}^{2} = {(\frac{- 3}{2})}^{2} + {(\frac{3}{2})}^{2} + {(- 3)}^{2} = \frac{9}{4} + \frac{9}{4} + 9 = 9 + \frac{9}{2} = \frac{27}{2}$

Hence, $2 P Q^{2} = 27$ .

Solution

1 25

2 19

3 27

4 29

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