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