Question

The image of the point (1,6,3) in the $\frac{x}{1} = \frac{y -1}{2} = \frac{z -2}{3}$ is

• A. (1,0,7)
• B. (7,0,1)
• C. (2,7,0)
• D. (-1,-6,-3)

Answer 1

Answer: (A)

$M \equiv (\lambda , 2 \lambda + 1, 3 \lambda + 2)$ is the point on the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2 }{3}\,\,\,\,\,\dots(1)$
$\therefore \, PM = (\lambda - 1 , 2\lambda - 5 , 3 \lambda - 1)$
PM is perpendicular to the line (1)
$\Rightarrow \,1(\lambda - 1) + 2(2\lambda - 5) + 3(3\lambda - 1) = 0$
$\Rightarrow \, \lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$
$\Rightarrow \, 14\lambda = 14 $
$ \Rightarrow \, \lambda = 1$
$\therefore \, M \equiv (1, 3, 5)$
M is the midpoint of PQ
$\Rightarrow \left(1,3,5\right) = \left(\frac{a+1}{2} , \frac{6+b}{2} , \frac{3+c}{2}\right)$
$ \Rightarrow \, a = 1, b = 0, c = 7 $
$\therefore \, Q = (1, 0 ,7)$ is the image of P