The reflection of the point P1,0,0 in the line x−12=y+1−3=z+108 is

Coordinates of any point Q on the given line are 2r+1,−3r−1,8r−10for some r∈RSo, the direction ratios of PQ are 2r,−3r−1,−8r−10Now PQ is perpendicular to the given line if 22r−3−3r−1+88r−10=0⇒77r−77=0⇒r=1and the coordinates of Q, the foot of the perpendicular from P on the line are (3, -4, -2).Let R (a, b, c) be the reflection of P in the given lines when Q is the mid-point of PR ⇒a+12=3,b2=−4,c2=−2⇒a=5,b=−8,c=−4and the coordinates of the required point are 5,−8,−4.

The reflection of the point P1,0,0 in the line x−12=y+1−3=z+108 is

Coordinates of any point Q on the given line are 2r+1,−3r−1,8r−10for some r∈RSo, the direction ratios of PQ are 2r,−3r−1,−8r−10Now PQ is perpendicular to the given line if 22r−3−3r−1+88r−10=0⇒77r−77=0⇒r=1and the coordinates of Q, the foot of the perpendicular from P on the line are (3, -4, -2).Let R (a, b, c) be the reflection of P in the given lines when Q is the mid-point of PR ⇒a+12=3,b2=−4,c2=−2⇒a=5,b=−8,c=−4and the coordinates of the required point are 5,−8,−4.

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The reflection of the point $P (1, 0, 0)$ in the line $\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}$ is

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(3, -4, -2)

(5, -8, -4)

(1, -1, -4)

(2, -3, 8)

answer is B.

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Detailed Solution

Coordinates of any point Q on the given line are $(2 r + 1, - 3 r - 1, 8 r - 10)$
for some $r \in R$
So, the direction ratios of $P Q$ are $2 r, - 3 r - 1, - 8 r - 10$
Now $P Q$ is perpendicular to the given line
if $2 (2 r) - 3 (- 3 r - 1) + 8 (8 r - 10) = 0$
$\Rightarrow 77 r - 77 = 0 \Rightarrow r = 1$
and the coordinates of Q, the foot of the perpendicular from P on the line
are (3, -4, -2).
Let R (a, b, c) be the reflection of P in the given lines when Q is the mid-point of PR
$\Rightarrow \frac{a + 1}{2} = 3, \frac{b}{2} = - 4, \frac{c}{2} = - 2$
$\Rightarrow a = 5, b = - 8, c = - 4$
and the coordinates of the required point are $(5, - 8, - 4)$ .