QNumerical Answer
If the image of the point $P(a, 2, a)$ in the line $\frac{x}{2} = \frac{y+a}{1} = \frac{z}{1}$ is $Q$ and the image of $Q$ in the line $\frac{x-2b}{2} = \frac{y-a}{1} = \frac{z+2b}{-5}$ is $P$ , then $a + b$ is equal to ________ .
1. Image of a Point in a Line (3D)
To find the image of a point $P$ in a line $L$, we first find the foot of the perpendicular $F$ from $P$ to $L$. If the line is given by $\vec{r} = \vec{a} + \lambda \vec{d}$, we find $\lambda$ by setting $(\vec{P} – (\vec{a} + \lambda \vec{d})) \cdot \vec{d} = 0$. Once $F$ is found, the image $Q$ is calculated using the midpoint property: $\vec{F} = \frac{\vec{P} + \vec{Q}}{2}$, which yields $\vec{Q} = 2\vec{F} – \vec{P}$. This is a standard procedure in JEE geometry.
2. Properties of Perpendicular Reflection
Reflection in a line ensures two critical geometric conditions: first, the segment connecting the point and its image is bisected by the line (the midpoint lies on the line). Second, the segment is perpendicular to the direction vector of the line. In this problem, the mutual reflection (P to Q and Q back to P) implies that both lines share the same perpendicular bisector plane for the segment PQ, even if the lines themselves are different.
3. Vector Dot Product in 3D Space
The dot product of two vectors $\vec{u}(x_1, y_1, z_1)$ and $\vec{v}(x_2, y_2, z_2)$ is defined as $x_1x_2 + y_1y_2 + z_1z_2$. In 3D geometry problems, the dot product is the primary tool used to enforce perpendicularity. If a line has direction ratios $(l, m, n)$ and a vector joining two points has components $(a, b, c)$, then $al + bm + cn = 0$ if and only if the vector is perpendicular to the line.
4. Parametric Form of a Line
A line expressed in the form $\frac{x-x_1}{l} = \frac{y-y_1}{m} = \frac{z-z_1}{n}$ is in its symmetric form. To perform calculations, it is often converted to parametric form: $x = x_1 + lr, y = y_1 + mr, z = z_1 + nr$. This allows us to represent any general point on the line using a single variable $r$, which can then be solved using auxiliary conditions like distance or perpendicularity.
1
What is the significance of $a, b \in \mathbb{R}$?
In this problem, $a$ and $b$ are parameters that define the coordinates and the lines. The problem implies they are real numbers, and we solve for them using geometric constraints.
2
Can P and Q be the same point?
If $P$ and $Q$ were the same, $P$ would have to lie on the line. However, the conditions given usually lead to distinct points unless specified otherwise.
3
Why did we use the midpoint formula?
By definition of reflection (image), the line of reflection acts as a mirror, placing the image at the same distance on the opposite side, making the intersection point the midpoint.
4
Is the direction vector $(2, 1, 1)$?
Yes, for the first line, the denominators in the symmetric form represent the direction ratios, which form the direction vector $(2, 1, 1)$.
5
What is the image of $(1, 2, 3)$ in the x-axis?
In the x-axis, the x-coordinate stays the same and others negate, so $(1, -2, -3)$. This is a simpler version of the line reflection used here.
6
How do I verify the value of $a=1$?
Substitute $a=1$ into the dot product equation $-12a + 12 = 0$. Since it satisfies it, the perpendicularity condition is met.
7
What if the lines were parallel?
If the lines were parallel, the segments $PQ$ for both reflections would be parallel, but since the direction vectors $(2,1,1)$ and $(2,1,-5)$ are not proportional, the lines are not parallel.
8
Can this be solved using planes?
Yes, you could define the plane perpendicular to the line passing through $P$, find the intersection (foot), and then find the image. It’s essentially the same logic.
9
Is this a 2026 JEE Main question?
Yes, the metadata from the source image indicates it is Question 71 from a 2026 JEE Main session.
10
What is the most common mistake here?
Calculating the foot $F$ but forgetting to find the image $Q$ (i.e., stopping at $Q=F$) or making sign errors in the dot product expansion.