If the image of the pointP(a, 2, a)$P(a, 2, a)$ in the line (x)/(2) = (y + a)/(1) = (z)/(1)$\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1}$ is Q$Q$ and the image of Q$Q$ in the line (x - 2b)/(2) = (y - a)/(1) = (z + 2b)/(-5)$\frac{x - 2b}{2} = \frac{y - a}{1} = \frac{z + 2b}{-5}$ is P$P$, then a + b$a + b$ is equal to
Numerical Answer Type:
Enter a numerical valueAnswer: 3 to 3+4 marks
Solution & Explanation
Related Formula
For a point and its image across a line, the midpoint of the segment connecting them lies exactly on the line. The segment itself is perpendicular to the line.
Core Logic
Line in 3D diagram for Q22 - JEE Main 2026 Evening
Let the first line be L₁$L_1$: (x)/(2) = (y + a)/(1) = (z)/(1) = λ$\frac{x}{2} = \frac{y + a}{1} = \frac{z}{1} = \lambda$.
A general point on L₁$L_1$ is (2λ, λ - a, λ)$(2\lambda, \lambda - a, \lambda)$. This is the midpoint M₁$M_1$ of P(a, 2, a)$P(a, 2, a)$ and Q(x₁, y₁, z₁)$Q(x_1, y_1, z_1)$.
Now, the image of Q$Q$ across L₂$L_2$: (x - 2b)/(2) = (y - a)/(1) = (z + 2b)/(-5) = μ$\frac{x - 2b}{2} = \frac{y - a}{1} = \frac{z + 2b}{-5} = \mu$ is P(a, 2, a)$P(a, 2, a)$.
The midpoint M₂$M_2$ of PQ$PQ$ lies on L₂$L_2$. But M₂$M_2$ is the exact same set of coordinates in space as M₁$M_1$ since they are both the midpoint of the segment PQ$PQ$!
Step 1: Equating Midpoints
General point on L₂$L_2$ is (2μ + 2b, μ + a, -5μ - 2b)$(2\mu + 2b, \mu + a, -5\mu - 2b)$.
Since the midpoint must be the same:
(a + (4λ - a))/(2) = 2μ + 2b 2λ = 2μ + 2b λ - μ = b$\frac{a + (4\lambda - a)}{2} = 2\mu + 2b \implies 2\lambda = 2\mu + 2b \implies \lambda - \mu = b$(2 + (2λ - 2a - 2))/(2) = μ + a λ - a = μ + a λ - μ = 2a$\frac{2 + (2\lambda - 2a - 2)}{2} = \mu + a \implies \lambda - a = \mu + a \implies \lambda - \mu = 2a$(a + (2λ - a))/(2) = -5μ - 2b λ = -5μ - 2b λ + 5μ = -2b$\frac{a + (2\lambda - a)}{2} = -5\mu - 2b \implies \lambda = -5\mu - 2b \implies \lambda + 5\mu = -2b$
Step 2: Solving System
From the first two equations, b = 2a$b = 2a$.
Substitute b = 2a$b = 2a$ into the third: λ + 5μ = -4a$\lambda + 5\mu = -4a$.
Also λ - μ = 2a$\lambda - \mu = 2a$. Subtracting gives 6μ = -6a μ = -a$6\mu = -6a \implies \mu = -a$.
Thus λ = a$\lambda = a$.
Line in 3D diagram for Q22 - JEE Main 2026 Evening
The direction vector of segment PQ$PQ$ is proportional to the vector from P$P$ to M₁$M_1$.
PM₁ = (2λ - a) i + (λ - a - 2) j + (λ - a) k$\vec{PM_1} = (2\lambda - a)\hat{i} + (\lambda - a - 2)\hat{j} + (\lambda - a)\hat{k}$.
Since λ = a$\lambda = a$:
PM₁ = a i - 2 j + 0 k$\vec{PM_1} = a\hat{i} - 2\hat{j} + 0\hat{k}$.
This vector must be perpendicular to L₁$L_1$ (direction vector 2 i + j + k$2\hat{i} + \hat{j} + \hat{k}$):
(a)(2) + (-2)(1) + (0)(1) = 0 2a - 2 = 0 a = 1$(a)(2) + (-2)(1) + (0)(1) = 0 \implies 2a - 2 = 0 \implies a = 1$.
Step 3: Final Values
Since a = 1$a = 1$, and b = 2a$b = 2a$, we have b = 2$b = 2$.
Therefore, a + b = 1 + 2 = 3$a + b = 1 + 2 = 3$.
Pattern Recognition
If P$P$ is the image of Q$Q$ on L₁$L_1$, and Q$Q$ is the image of P$P$ on L₂$L_2$, then L₁$L_1$ and L₂$L_2$ must intersect segment PQ$PQ$ at the exact same midpoint. Using the midpoint consistency across both lines creates a rapid linear system.
Keywords:#image of the point#JEE Main 2026 Evening Q22#Three Dimensional Geometry JEE Main 2026#Line in 3D JEE Main 2026
More Three Dimensional Geometry Previous-Year Questions — Page 10
Q29jee_main_2024_01_february_morningShortest Distance Between Skew Lines
Let the line of the shortest distance between the lines L₁: r=( i+2 j+3 k)+λ( i- j+ k)$L_{1}:\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and L₂: r=(4 i+5 j+6 k)+μ( i+ j- k)$L_{2}:\vec{r}=(4\hat{i}+5\hat{j}+6\hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})$ intersect L₁$L_{1}$ and L₂$L_{2}$ at P and Q respectively. If $(α, β, γ)$$$(\alpha, \beta, \gamma)$$is the midpoint of the line segment PQ, then$ is the midpoint of the line segment PQ, then $2(\alpha+\beta+\gamma)$ is equal to
Numerical Answer.Answer: 21 to 21
Solution
Related Formula
The vector connecting the shortest distance points P$P$ and Q$Q$ on two skew lines must be simultaneously perpendicular to the direction vectors b₁$\vec{b}_1$ and b₂$\vec{b}_2$ of both lines:
Angle Between Two Lines diagram for Q3 - JEE Main 2024 Evening
Since θ = (1)/(2)$\cos \theta = \frac{1}{2}$, we have:
θ = (π)/(3)$$\theta = \frac{\pi}{3}$$
Pattern Recognition
When asked for an angle like ∠ QPR$\angle QPR$, always ensure both vectors diverge from the common vertex P$P$ (i.e., use PQ$\overrightarrow{PQ}$ and PR$\overrightarrow{PR}$) to avoid sign errors.
Chapter Mix
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Q26jee_main_2024_29_january_eveningShortest Distance Between Two Lines
Let O$O$ be the origin, and M$M$ and N$N$ be the points on the lines (x - 5)/(4) = (y - 4)/(1) = (z - 5)/(3)$\frac{x - 5}{4} = \frac{y - 4}{1} = \frac{z - 5}{3}$ and (x + 8)/(12) = (y + 2)/(5) = (z + 11)/(9)$\frac{x + 8}{12} = \frac{y + 2}{5} = \frac{z + 11}{9}$ respectively such that MN$MN$ is the shortest distance between the given lines. Then OM· ON$\overline{\mathrm{OM}}\cdot \overline{\mathrm{ON}}$ is equal to
Numerical Answer.Answer: 9 to 9
Solution
Related Formula
The line segment of shortest distance MN$MN$ is perpendicular to both directing vectors b₁$\vec{b}_1$ and b₂$\vec{b}_2$.
Core Logic
Let general point coordinates be expressions of parameters λ$\lambda$ and μ$\mu$:
Shortest distance points constrain line segments to align with the direct cross product vector. Setting proportional components yields quick parameters.
Chapter Mix
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Q3jee_main_2024_27_jan_morningDistance of a Point from a Line
The distance of the point (7, -2, 11)$(7, -2, 11)$ from the line (x-6)/(1)=(y-4)/(0)=(z-8)/(3)$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line (x-5)/(2)=(y-1)/(-3)=(z-5)/(6)$\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is:
Let point A = (7, -2, 11)$A = (7, -2, 11)$. We want the distance from A$A$ to the line L₁: (x-6)/(1)=(y-4)/(0)=(z-8)/(3)$L_1: \frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ measured along a line parallel to L₂: (x-5)/(2)=(y-1)/(-3)=(z-5)/(6)$L_2: \frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$.
The line passing through A$A$ parallel to L₂$L_2$ will have the equation:
Distance of a point from a line along another direction means finding the intersection of the given line and a new line passing through the point parallel to the direction vector. The zero in the direction ratio is a massive shortcut—just equate the corresponding numerator to zero.
Chapter Mix
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Q8jee_main_2024_27_jan_morningShortest Distance Between Two Lines
If the shortest distance between the lines(x-4)/(1)=(y+1)/(2)=(z)/(-3)$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and (x-λ)/(2)=(y+1)/(4)=(z-2)/(-5)$\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is 6√(5)$\frac{6}{\sqrt{5}}$, then the sum of all possible values of λ$\lambda$ is:
Standard Shortest Distance methodology between skew lines. Cross product of direction vectors forms the perpendicular frame normal, and dot-producting the difference of positional anchor points yields the direct orthogonal projection.
Chapter Mix
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More Three Dimensional Geometry Questions — jee_main_2026_23_january_evening
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
