Question

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

Answer 1

The cartesian equation of the line is $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ ......$\left(1\right)$

The given point is $(1,6,3)$

To find the image of $P(1,6,3)$ in the line draw a line $PR$ is perpendicular to the line .

Let $R$ be the image of $P$ and $Q$ is the mid point of $PR$

Let $a,b,c$ be the direction cosines of $PR$.Since $PR$ is perpendicular to the line,

By applying the condition of perpendicularity.

$a\times 1+b\times 2+c\times 3=0$ -------$\left(2\right)$

$\Rightarrow a+2b+3c=0$

Let $\frac{x-1}{a}=\frac{y-6}{b}=\frac{z-3}{c}=k$(say) -------$\left(3\right)$

Any point on the line $\left(2\right)$ is $(ak+1,bk+6,ck+3)$

Let the point be $Q$

But $Q$ also lies on line $\left(1\right)$

$\therefore \frac{ak+1}{1}=\frac{bk+6-1}{2}=\frac{ck+3-2}{3}$

$\Rightarrow \frac{ak+1}{1}=\frac{bk+5}{2}=\frac{ck+1}{3}$

$\Rightarrow \frac{ak+1+2bk+10+3ck+3}{1\times 1+2\times 2+3\times 3}$

$\Rightarrow \frac{14+(a+2b+3c)k}{14}=1$

$\Rightarrow 14+(a+2b+3c)k=14$

$\Rightarrow ak=0,bk=-3$ and $ck=2$

$Q=(0+1,-3+6,2+3)=(1,3,5)$

Since $Q$ is the midpoint of $PR$

$\frac{1+x^{\prime }}{2}=1$ and $\frac{6+y^{\prime }}{2}=3$ and $\frac{3+z^{\prime }}{2}=5$

Hence,$x^{\prime }=1$,$y^{\prime }=0$ and $z^{\prime }=7$

Hence $(1,0,7)$ is the image of $P$ in line $\left(1\right)$