Let and be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, –1) and , then the value of is __________. [2025]

Question

Let $L_{1} : \frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0}$ and $L_{2} : \frac{x - 2}{2} = \frac{y}{0} = \frac{z + 4}{α}, α \in R$ be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, –1) and $L_{2}$ , then the value of $26 α {(P B)}^{2}$ is __________. [2025]

Smart Achievers · Accepted Answer

(216)

Point $B \equiv (3 μ + 1, - μ + 1, - 1) \equiv (2 λ + 2, 0, α λ - 4)$

$\Rightarrow 3 μ + 1 = 2 λ + 2, - μ + λ = 0, λ α - 4 = - 1$

$∴ μ = 1 \Rightarrow λ = 1 and α = 3$

So, point B(4, 0, –1).

Let point P is (2k + 2, 0, 3k – 4).

So, Dr's of AP is < 2k + 1, –1, 3k – 3 >

Since, $L_{2} ⊥ A P$

$\Rightarrow 2 (2 k + 1) + 0 (- 1) + 3 (3 k - 3) = 0$

$\Rightarrow 4 k + 2 + 9 k - 9 = 0 \Rightarrow k = \frac{7}{13}$

$∴ Point P \equiv (\frac{40}{13}, 0, \frac{- 31}{13})$

$∴ {(P B)}^{2} = {(\frac{40}{13} - 4)}^{2} + {(0 - 0)}^{2} + {(\frac{- 31}{13} + 1)}^{2} = \frac{468}{169}$

$∴ 26 α {(P B)}^{2} = 26 \times 3 \times \frac{468}{169} = 216$ .

Solution

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Solution

Q. Let L1:x–13=y–1–1=z+10 and L2:x–22=y0=z+4α, α∈R be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, –1) and L2, then the value of 26α(PB)2 is __________. [2025]

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