Q.(b) Find the image of the point $(1, 2, 1)$ with respect to the line $\dfrac{x-3}{1} = \dfrac{y+1}{2} = \dfrac{z-1}{3}$. Also find the equation of the line joining the given point and its image.

Concept: Image of a point in a line — the line is the perpendicular bisector of the segment joining the point and its image.

Step 1: Let the given point be $P(1,2,1)$. The line is $\frac{x-3}{1} = \frac{y+1}{2} = \frac{z-1}{3} = \lambda$.

A general point on the line is $Q(3+\lambda, -1+2\lambda, 1+3\lambda)$.

Step 2: The direction ratios of $PQ$ are $(2+\lambda, -3+2\lambda, 3\lambda)$. For $Q$ to be the foot of the perpendicular, $PQ$ must be perpendicular to the line’s direction vector $(1,2,3)$:

Step 3: Foot of perpendicular $M = \left(3+\frac{2}{7}, -1+\frac{4}{7}, 1+\frac{6}{7}\right) = \left(\frac{23}{7}, -\frac{3}{7}, \frac{13}{7}\right)$.

If $P'(x,y,z)$ is the image, then $M$ is the midpoint of $PP'$:

Step 4: Equation of line joining $P(1,2,1)$ and $P'\left(\frac{39}{7}, -\frac{20}{7}, \frac{19}{7}\right)$ has direction ratios $\left(\frac{32}{7}, -\frac{34}{7}, \frac{12}{7}\right) \propto (16, -17, 6)$.

Concept and Intuition

The image of a point in a line is the mirror reflection across that line. Think of the line as a mirror in 3D space. The key idea: the foot of the perpendicular from the point to the line is the midpoint of the point and its image. Why? Because reflection preserves distances and the line of reflection is the perpendicular bisector of the segment joining a point and its image.

So the plan is:

1. Find the foot of the perpendicular from $P$ to the given line.
2. Use the midpoint formula to get the image $Q$.
3. Write the equation of line $PQ$.

Step-by-step solution

1. Parameterize the given line

The line is $\frac{x-3}{1} = \frac{y+1}{2} = \frac{z-1}{3} = \lambda$ (say).

Then any point on the line is:

2. Find the foot of the perpendicular from $P$ to the line

Let $F$ be the foot. Then $F = (3 + \lambda, -1 + 2\lambda, 1 + 3\lambda)$ for some $\lambda$.

The vector $\overrightarrow{PF}$ is:

The direction vector of the line is $\vec{d} = (1, 2, 3)$.

Since $PF$ is perpendicular to the line, $\overrightarrow{PF} \cdot \vec{d} = 0$:

So the foot $F$ is:

3. Find the image $Q$ using the midpoint property

$F$ is the midpoint of $P(1,2,1)$ and $Q(x,y,z)$:

Solving:

So the image is $Q\left(\frac{39}{7}, -\frac{20}{7}, \frac{19}{7}\right)$.

4. Equation of the line joining $P$ and $Q$

The direction vector of $PQ$ is:

Multiply by 7 to get integer direction: $(32, -34, 12)$. Divide by 2 for simplicity: $(16, -17, 6)$.

Using point $P(1,2,1)$, the line equation in symmetric form is: