The foot of perpendicular of the point (2, 0, 5) on the line x+12=y−15=z+1−1 is (α,β,γ) Then whichof the following is NOT correct?

L≅x+12=y−15=z+1−1(=λ)Q=(x,y,z)=(2λ−1,5λ+1,−λ−1)be the foot of perpendicular of P2,0,5Dr'⁡sPQ→=(2λ−3,5λ+1,−λ−6)PQ→⊥L⇒2(2λ−3)+5(5λ+1)−1(−λ−6)=04λ−6+25λ+5+λ+6=030λ=−5,λ=−16Q=−26−1,−56+1,16−1Q=−43,16,−561. αβγ=−43λ1r−5t=1415 correct 2. βγ=16−56=−15 incorect givne βγ=−53. γα=−56−43=56×34=58 correct

The foot of perpendicular of the point (2, 0, 5) on the line x+12=y−15=z+1−1 is (α,β,γ) Then whichof the following is NOT correct?

L≅x+12=y−15=z+1−1(=λ)Q=(x,y,z)=(2λ−1,5λ+1,−λ−1)be the foot of perpendicular of P2,0,5Dr'⁡sPQ→=(2λ−3,5λ+1,−λ−6)PQ→⊥L⇒2(2λ−3)+5(5λ+1)−1(−λ−6)=04λ−6+25λ+5+λ+6=030λ=−5,λ=−16Q=−26−1,−56+1,16−1Q=−43,16,−561. αβγ=−43λ1r−5t=1415 correct 2. βγ=16−56=−15 incorect givne βγ=−53. γα=−56−43=56×34=58 correct

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The foot of perpendicular of the point (2, 0, 5) on the line $\frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{- 1} is (α, β, γ)$ Then which
of the following is NOT correct?

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$\frac{α β}{γ} = \frac{4}{15}$

$\frac{α}{β} = - 8$

$\frac{β}{γ} = - 5$

$\frac{γ}{α} = - \frac{5}{8}$

answer is B.

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Detailed Solution

$L ≅ \frac{x + 1}{2} = \frac{y - 1}{5} = \frac{z + 1}{- 1} (= λ)$

$Q = (x, y, z) = (2 λ - 1, 5 λ + 1, - λ - 1) b e t h e f o o t o f p e r p e n d i c u l a r o f P (2, 0, 5)$

${Dr}^{'} s \vec{PQ} = (2 λ - 3, 5 λ + 1, - λ - 6)$

$\vec{PQ} ⊥ L \Rightarrow 2 (2 λ - 3) + 5 (5 λ + 1) - 1 (- λ - 6) = 0$

$\begin{array}{l} 4 λ - 6 + 25 λ + 5 + λ + 6 = 0 \end{array}$
$\begin{array}{l} 30 λ = - 5, λ = - \frac{1}{6} \\ Q = (\frac{- 2}{6} - 1, \frac{- 5}{6} + 1, \frac{1}{6} - 1) \\ Q = (\frac{- 4}{3}, \frac{1}{6}, \frac{- 5}{6}) \end{array}$
$1 . \frac{α β}{γ} = \frac{\frac{- 4}{3} \frac{λ 1}{r}}{\frac{- 5}{t}} = \frac{14}{15} correct$
$2. \frac{β}{γ} = \frac{\frac{1}{6}}{\frac{- 5}{6}} = \frac{- 1}{5} incorect givne \frac{β}{γ} = - 5$
$3 . \frac{γ}{α} = \frac{\frac{- 5}{6}}{\frac{- 4}{3}} = \frac{5}{6} \times \frac{3}{4} = \frac{5}{8} correct$