Two lines (x-3/1) = (y+1/3) = (z-6/-1) and (x+5/7) = (y-2/-6) = (z-3/4) intersect at the point R. The reflection of R in the xy-plane has coordinates :

Question

Two lines (x-3/1) = (y+1/3) = (z-6/-1) and (x+5/7) = (y-2/-6) = (z-3/4) intersect at the point R. The reflection of R in the xy-plane has coordinates : (A) (2, 4, 7) (B) (-2, 4, 7) (C) (2, -4, -7) (D) (2, -4, 7)

Tardigrade Admin · Accepted Answer

Point on L1 (λ + 3, 3λ - 1, - λ + 6) Point on L2 (7 μ - 5, - 6 μ + 2, 4μ + 3) ⇒ λ + 3 = 7 μ - 5 ...(i) 3 λ - 1 = - 6 μ + 2 ...(ii) ⇒ λ = -1, μ = 1 point R(2,-4,7) Reflection is (2,-4,-7)

Q. Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{- 1}$ and $\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4}$ intersect at the point $R$ . The reflection of $R$ in the xy-plane has coordinates :

$(2, 4, 7)$

$(- 2, 4, 7)$

$(2, - 4, - 7)$

$(2, - 4, 7)$

Point on $L_{1} (λ + 3, 3 λ - 1, - λ + 6)$
Point on $L_{2} (7 μ - 5, - 6 μ + 2, 4 μ + 3)$
$\Rightarrow λ + 3 = 7 μ - 5$ ...(i)
$3 λ - 1 = - 6 μ + 2$ ...(ii)
$\Rightarrow λ = - 1, μ = 1$
point $R (2, - 4, 7)$
Reflection is $(2, - 4, - 7)$

Q. Two lines 1x−3​=3y+1​=−1z−6​ and 7x+5​=−6y−2​=4z−3​ intersect at the point R. The reflection of R in the xy-plane has coordinates :

(2,4,7)

(−2,4,7)

(2,−4,−7)

(2,−4,7)