Question

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

• A. $(1,0,7)$
• B. $(1,3,5)$
• C. $(2,4,6)$
• D. $(-1,-6,-3)$

Answer 1

The correct option is A $(1,0,7)$

Let $P$ be the given point and let $L$ be the foot of perpendicular from $P$ to the given line.

The coordinates of a general point on the given line are given by
$\frac{x-0}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$
i.e.,

$x=\lambda ,y=2\lambda +1,z=3\lambda +2$

Let the coordinates of $L$ be
$(\lambda ,2\lambda +1,3\lambda +2)$
So, direction ratios of $PL$ are
$(\lambda -1,2\lambda -5,3\lambda -1)$

Direction ratios of the given lines are $(1,2,3)$ which is perpendicular to $PL$

Therefore,
$(\lambda -1)\cdot 1+(2\lambda -5)\cdot 2+(3\lambda -1)\cdot 3=0$
$\Rightarrow \lambda =1$

So, coordinates of $L$ are $(1,3,5)$.
Let $Q(x_1,y_1,z_1)$ be the image of $P(1,6,3)$ on given line.
Where, $L$ is mid-point of $PQ$.

$\therefore 1=\frac{x_1+1}{2},3=\frac{y_1+6}{2},5=\frac{z_1+3}{2}$

$\Rightarrow x_1=1,y_1=0,z_1=7$

$\therefore$ Image of $P(1,6,3)$ in the given line is $(1,0,7)$.

Hence, option A.