How do I find the image of a point in a line in 3D space?

Question

The question is,

find the image of the point $(1, 6, 3)$ in the line $$\frac x1 = \frac {y-1}{2} = \frac {z-2}{3}$$

I want to know the general equation to find the image of a point in a line.

EDIT

By image, I mean a point which is diametrically opposite to the point, on the opposite side of the line

Answer 1

Let $Q(a,b,c)$ be the image of $P(1,6,3)$ in the given line.Then direction ratios of $PQ$ are $(a-1,b-6,c-3)$
As the given line and $PQ$ will be perpendicular to each other.So $a_1a_2+b_1b_2+c_1c_2=0$ is the condition of perpendicularity
$1(a-1)+2(b-6)+3(c-3)=0$
$a+2b+3c=22..........(1)$
Now the mid point of $PQ(\frac{a+1}{2},\frac{b+6}{2},\frac{c+3}{2})$ will lie on the line.
Write the equation of line in parametric form
$x=t,y=2t+1,z=3t+2$

So $\frac{a+1}{2}=t,\frac{b+6}{2}=2t+1,\frac{c+3}{2}=3t+2$
$a=2t-1,b=4t-4,c=6t+1$

Put this in equation $(1)$,find $t$ and hence find $(a,b,c)$

Answer 2

HINT: set $$x=t$$, $$\frac{y-1}{2}=t$$ and $$\frac{z-2}{3}=t$$ and you will get the equation of the line

Answer 3

Write everything in terms of z. $x=\frac{z-2}{3},y=\frac{2z-1}{3},z=z$. Then convert to vector form. $\left( \begin{array}{c} x \\ y \\ z \\ \end{array} \right)=z \left( \begin{array}{c} \frac{1}{3} \\ \frac{2}{3} \\ 1 \\ \end{array} \right)+\left( \begin{array}{c} -\frac{2}{3} \\ -\frac{1}{3} \\ 1 \\ \end{array} \right)$

Consider this as the vector equation $\vec{x}=z\vec{v}+\vec{p}$. Now project the point $\vec{u}=(1,6,3)$ onto this line with $\vec{w}=\frac{\vec{u}.\vec{v}}{\vec{v}.\vec{v}}\vec{v}$ giving $(1,3,5)$

Answer 4

To find the image of a point in a line, you must first find the equation of the normal to the line through the point.

Find $\vec m $ so that $\vec l . \vec m = 0$ where $\vec l$ is the vector of the line and $\vec m$ is the vector of the perpendicular through the point.

find the coords of the point of intersection of the two lines.

The image of a point from a line is the same distance from the line as the point itself. So, we can use the mid-point formula of a line segment, solving for $x', y',$ and $z'$ where $x', y',$ and $z'$ are coords of the image.