A line L passing through the point P (2,4,3) is perpendicular to both the lines (x-1/2)=(y+3/1)=(z-2/4) and (x-2/3)=(y+1/2)=(1-z/2). If the position vector of point Q on L is (a, b, c) such that (P Q)2=357 then a + b + c can be

Question

A line L passing through the point P (2,4,3) is perpendicular to both the lines (x-1/2)=(y+3/1)=(z-2/4) and (x-2/3)=(y+1/2)=(1-z/2). If the position vector of point Q on L is (a, b, c) such that (P Q)2=357 then a + b + c can be (A) 16 (B) 15 (C) 2 (D) 1

Tardigrade Admin · Accepted Answer

Equation of L: (x-2/10)=(y-4/-16)=(z-3/-1)=λ PQ =10 λ hat i -16 λ hat j -λ hat k ⇒ 35 λ2=357 ⇒ λ= ± 1 Q=(2+10 λ, 4-16 λ, 3-λ)

Q. A line L passing through the point P(2,4,3) is perpendicular to both the lines 2x−1​=1y+3​=4z−2​ and 3x−2​=2y+1​=21−z​. If the position vector of point Q on L is (a,b,c) such that (PQ)2=357 then a+b+c can be

16

15

2

1

Equation of L:10x−2​=−16y−4​=−1z−3​=λ PQ​=10λi^−16λj^​−λk^ ⇒35λ2=357⇒λ=±1 Q=(2+10λ,4−16λ,3−λ)

Equation of $L : \frac{x - 2}{10} = \frac{y - 4}{- 16} = \frac{z - 3}{- 1} = λ$
$PQ = 10 λ \hat{i} - 16 λ \hat{j} - λ \hat{k}$
$\Rightarrow 35 λ^{2} = 357 \Rightarrow λ = \pm 1$
$Q = (2 + 10 λ, 4 - 16 λ, 3 - λ)$