Let the lines and intersect at the point R. Let P and Q be the points lying on lines and respectively, such that and . If the point P lies in the first octant, then is equal to [2026]

Question

Let the lines $L_{1} : \vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k}), λ \in ℝ$ and $L_{2} : \vec{r} = (4 \hat{i} + \hat{j}) + μ (5 \hat{i} + 2 \hat{j} + \hat{k}), μ \in ℝ,$ intersect at the point R. Let P and Q be the points lying on lines $L_{1}$ and $L_{2}$ respectively, such that $| \vec{P R} |$ $= \sqrt{29}$ and $| \vec{P Q} |$ $= \sqrt{\frac{47}{3}}$ . If the point P lies in the first octant, then $27 {(Q R)}^{2}$ is equal to [2026]

Smart Achievers · Accepted Answer

(3)

$For POI$

$2 λ + 1 = 5 μ + 4; 3 λ + 2 = 2 μ + 1; 4 λ + 3 = μ$

$\Rightarrow λ = μ = - 1$

$R (- 1, - 1, - 1), P (2 λ + 1, 3 λ + 2, 4 λ + 3)$

$P R^{2} = 29 \Rightarrow {(2 λ + 2)}^{2} + {(3 λ + 3)}^{2} + {(4 λ + 4)}^{2} = 29$

$\Rightarrow λ = 0 or λ = - 2 (Reject)$

$\Rightarrow P (1, 2, 3)$

$Q (5 μ + 4, 2 μ + 1, μ)$

$| P Q |$ $= \sqrt{\frac{47}{3}} \Rightarrow P Q^{2} = \frac{47}{3}$

$\Rightarrow {(5 μ + 3)}^{2} + {(2 μ - 1)}^{2} + {(μ - 3)}^{2} = \frac{47}{3}$

$\Rightarrow μ = - \frac{1}{3}$

$Q = (\frac{7}{3}, \frac{1}{3}, - \frac{1}{3})$

${(Q R)}^{2} = {(\frac{7}{3} + 1)}^{2} + {(\frac{1}{3} + 1)}^{2} + {(- \frac{1}{3} + 1)}^{2}$

$= \frac{100 + 16 + 4}{9} = \frac{120}{9}$

$\Rightarrow 27 \times {(Q R)}^{2} = 27 \times \frac{120}{9} = 360$

Solution

1 320

2 340

3 360

4 348

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