Three-Dimensional Geometry Previous Year Questions (JEE Main 2026)
Solved JEE Main 2026 previous year questions on Three-Dimensional Geometry with concise, step-by-step solutions covering lines, shortest distance, image of a point, and foot of perpendicular.
Solved JEE Main 2026 previous year questions from Three-Dimensional Geometry, each worked out step by step.
Solutions are AI-generated and pending review.
Solution
Let $P = (2s, 3s, 3s-1)$ on the first line and $Q = (-1-t, 2+t, 4t)$ on the second.
The vector $\vec{PQ} = Q - P$ must be parallel to $(1, -1, 2)$:
$$\vec{PQ} = (-1-t-2s,\ 2+t-3s,\ 4t-3s+1)$$Setting $\vec{PQ} \parallel (1,-1,2)$ gives $s = \tfrac{1}{5},\ t = -\tfrac{8}{15}$.
Then $P = \left(\tfrac{2}{5}, \tfrac{3}{5}, -\tfrac{2}{5}\right)$ and $Q = \left(-\tfrac{7}{15}, \tfrac{22}{15}, -\tfrac{32}{15}\right)$.
$$\alpha^2 = |\vec{PQ}|^2 = \frac{338}{75} \implies 225\alpha^2 = 225 \cdot \frac{338}{75} = 3 \cdot 338 = 1014$$Answer: B (1014)
Solution
Move from $(-2,-8,6)$ along direction $(1,-1,2)$: the moving point is
$$X = (-2+r,\ -8-r,\ 6+2r)$$This must meet the target line $(1+u,\ 1+2u,\ -u)$. Equating components:
$$-2+r = 1+u,\quad -8-r = 1+2u,\quad 6+2r = -u$$Solving gives $r = -1,\ u = -4$, so the meeting point is $X = (-3, -7, 4)$.
$$\text{distance}^2 = (-3+2)^2 + (-7+8)^2 + (4-6)^2 = 1 + 1 + 4 = 6$$Answer: B (6)
Solution
The foot $M$ is the midpoint of $P$ and its image, and $M$ lies on $L$ with $\vec{PM}\perp(1,2,b)$. Writing $M$ on $L$ as $(0,1,a-1)+u(1,2,b)$ and imposing the perpendicularity and midpoint conditions gives two families; the constraint $b>0$ selects
$$a = 3,\quad b = 3,\quad c = 4,\quad u = 1$$So $L : (0,1,2) + u(1,2,3)$ with $|(1,2,3)| = \sqrt{14}$, and the foot is $M = (1,3,5)$.
For $S$ at distance $2\sqrt{14}$ from $M$: $|u_S - 1|\sqrt{14} = 2\sqrt{14} \implies u_S = 3$ or $u_S = -1$.
$$u_S = 3 \Rightarrow S = (3, 7, 11)\ (\alpha>0);\qquad u_S = -1 \Rightarrow S = (-1,-1,-1)$$Taking $\alpha > 0$: $S = (3,7,11)$, so $\alpha + \beta + \gamma = 21$.
Answer: C (21)
Solution
Direction of $L$ is $\vec{d_1}\times\vec{d_2}$ with $\vec{d_1}=(3,5,7)$, $\vec{d_2}=(1,4,7)$:
$$\vec{d} = (3,5,7)\times(1,4,7) = (7,\ -14,\ 7) \parallel (1,-2,1)$$With $L_3$ direction $\vec{d_3} = (2,1,2)$:
$$\cos\theta = \frac{|(1)(2)+(-2)(1)+(1)(2)|}{\sqrt{6}\,\sqrt{9}} = \frac{2}{3\sqrt{6}}$$$$\sin\theta = \sqrt{1 - \frac{4}{54}} = \sqrt{\frac{50}{54}} = \frac{5\sqrt{2}}{3\sqrt{6}}$$$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{5\sqrt{2}}{2} = \frac{5}{2}\sqrt{2}$$Answer: B ($\frac{5}{2}\sqrt{2}$)
Solution
Point on line 1: $(-1+3s,\ -a+5s,\ -b-1+7s)$. Point on line 2: $(2+t,\ b+4t,\ 2a+7t)$.
The intersection lies on the $xy$-plane, so its $z$-coordinate is $0$: $-b-1+7s = 0$.
Equating all three coordinates plus this condition and solving:
$$a = 3,\quad b = 4,\quad s = \tfrac{5}{7},\quad t = -\tfrac{6}{7}$$$$a + b = 3 + 4 = 7$$Answer: C (7)
Solution
Direction of $L$ is $(2,2,1)\times(1,2,2) = (2,\ -3,\ 2)$.
So $L : (1,1,1) + u(2,-3,2)$. Foot of perpendicular from $O$ satisfies $\vec{OP}\cdot(2,-3,2)=0$:
$$(1+2u)(2) + (1-3u)(-3) + (1+2u)(2) = 0 \implies 1 + 17u = 0 \implies u = -\tfrac{1}{17}$$$$P = \left(\tfrac{15}{17},\ \tfrac{20}{17},\ \tfrac{15}{17}\right) \implies a+b+c = \tfrac{50}{17}$$$$34(a+b+c) = 34 \cdot \tfrac{50}{17} = 100$$Answer: C (100)
Solution
With $\vec{d_1}=(2,-5,6)$, $\vec{d_2}=(0,1,-1)$:
$$\vec{n} = \vec{d_1}\times\vec{d_2} = (-1,\ 2,\ 2),\qquad |\vec{n}| = 3$$Separation of base points: $\vec{a_2}-\vec{a_1} = \left(-1,\ -2,\ -3\right)$.
$$\text{SD} = \frac{|(\vec{a_2}-\vec{a_1})\cdot\vec{n}|}{|\vec{n}|} = \frac{|(-1)(-1)+(-2)(2)+(-3)(2)|}{3} = \frac{|1-4-6|}{3} = \frac{9}{3} = 3$$Answer: B (3)
Solution
Two conditions must hold for a true image: the join is perpendicular to the line direction $(3,2,1)$, and the midpoint lies on the line.
Perpendicularity: $(\text{image}-\text{point})\cdot(3,2,1)=0$ gives $\alpha = 3$ or $\alpha = \tfrac{1}{4}$.
Midpoint on line: the midpoint must satisfy $\frac{x-2}{3}=\frac{y-1}{2}=\frac{z}{1}$. Testing:
- $\alpha = 3$: midpoint $(5,3,1)$ satisfies the line. ✓
- $\alpha = \tfrac{1}{4}$: midpoint $\left(\tfrac{7}{8},-\tfrac{3}{32},\tfrac{5}{16}\right)$ does not lie on the line. ✗
Only $\alpha = 3$ satisfies both conditions.
Answer: A (Only 3)
Solution
Let $A = (1+2t,\ 2+t,\ \alpha+3t)$ be the foot, with $\vec{PA}\perp(2,1,3)$.
Midpoint of $PA$ is $\left(0,\tfrac{3}{4},-\tfrac{1}{4}\right)$:
$$\frac{a+1+2t}{2}=0,\quad \frac{b+2+t}{2}=\frac34,\quad \frac{0+\alpha+3t}{2}=-\frac14$$together with $(A-P)\cdot(2,1,3)=0$. Solving:
$$a = 0,\quad b = 0,\quad \alpha = 1,\quad t = -\tfrac{1}{2}$$$$a^2 + b^2 + \alpha^2 = 0 + 0 + 1 = 1$$Answer: A (1)
Solution
With $\vec{d_1}=(1,2,-3)$, $\vec{d_2}=(2,4,-5)$:
$$\vec{n} = \vec{d_1}\times\vec{d_2} = (2,\ -1,\ 0),\qquad |\vec{n}| = \sqrt{5}$$Separation of base points: $\vec{a_2}-\vec{a_1} = (-6,\ 3,\ 3)$.
$$\text{SD} = \frac{|(-6)(2)+(3)(-1)+(3)(0)|}{\sqrt5} = \frac{|-15|}{\sqrt5} = \frac{15}{\sqrt5} = 3\sqrt{5}$$Answer: C ($3\sqrt{5}$)
Solution
Image $R$ of $P(0,-5,0)$ in line through $(1,0,-1)$, direction $(2,1,-2)$: the foot gives $R = 2F - P = (-2,\ 3,\ 2)$.
Image $S$ of $Q\left(0,-\tfrac12,0\right)$ in line through $(1,-9,-1)$, direction $(-1,4,1)$: $S = 2F - Q = \left(-2,\ -\tfrac{3}{2},\ 2\right)$.
Check $PQRS$ is a parallelogram: $\vec{Q}-\vec{P} = (0,\tfrac92,0) = \vec{R}-\vec{S}$. ✓
Area via diagonals $\vec{PR}$ and $\vec{QS}$:
$$\vec{PR} = (-2,\ 8,\ 2),\quad \vec{QS} = (-2,\ -1,\ 2)$$$$\vec{PR}\times\vec{QS} = (18,\ 0,\ 18),\qquad \text{Area} = \tfrac12|{\vec{PR}\times\vec{QS}}| = \tfrac12\cdot 18\sqrt2 = 9\sqrt2$$$$\text{Area}^2 = 162$$Answer: 162
Solution
Base point $A = (2,5,2)$, direction $\vec{d} = (2,3,4)$, $|\vec{d}|^2 = 29$.
$$\vec{AP} = (3,\ 1,\ 5),\qquad \vec{AP}\times\vec{d} = (-11,\ -2,\ 7)$$$$\text{dist}^2 = \frac{|\vec{AP}\times\vec{d}|^2}{|\vec{d}|^2} = \frac{121+4+49}{29} = \frac{174}{29} = 6$$Answer: C (6)
Solution
Line 1: $(1+a\lambda,\ 1-\lambda,\ -1)$. Line 2: $(4+2\mu,\ 0,\ -1+a\mu)$.
Equating $y$: $1-\lambda = 0 \Rightarrow \lambda = 1$. Equating $z$: $-1 = -1+a\mu \Rightarrow a\mu = 0$, and since $a\neq0$, $\mu = 0$.
Equating $x$: $1+a = 4 \Rightarrow a = 3$ (consistent with $a\neq0$).
Intersection point: $(4,\ 0,\ -1)$.
$$\text{dist}^2 = 4^2 + 0^2 + (-1)^2 = 16 + 1 = 17$$Answer: C (17)
Solution
Direction $(8,2,2)$, base $(-3,4,-1)$. Foot of perpendicular $F$ from $R(1,2,3)$:
$$F = (-3,4,-1) + t(8,2,2),\quad (F-R)\cdot(8,2,2)=0 \implies t = \tfrac12,\ F = (1,5,0)$$Since $P,Q$ are equidistant ($=6$) from $R$ on the line, they are symmetric about $F$, so $P+Q = 2F$. (Note $|RF| = 3\sqrt2 < 6$, so $P,Q$ exist.)
$$\text{Centroid} = \frac{P+Q+R}{3} = \frac{2F+R}{3} = \frac{(2,10,0)+(1,2,3)}{3} = (1,4,1)$$$$\alpha+\beta+\gamma = 1+4+1 = 6$$Answer: C (6)
Solution
Line through $(0,1,2)$, direction $(1,1,2)$. Foot of perpendicular from $(1,2,7)$:
$$(F-P)\cdot(1,1,2)=0 \implies F = (2,3,6),\qquad \text{image} = 2F-P = (3,4,5)$$Distance from $(a,2,5)$ to $(3,4,5)$ equals $4$:
$$(a-3)^2 + (2-4)^2 + 0 = 16 \implies (a-3)^2 = 12$$$$a = 3 \pm 2\sqrt3 \implies \text{sum} = 6$$Answer: C (6)
Solution
The foot $(1,\mu,2)$ lies on the line $(4,9,5)+t(1,2,1)$: $1 = 4+t \Rightarrow t = -3$, giving $\mu = 9+2(-3) = 3$ and $z = 5-3 = 2$. ✓
Perpendicularity $(\text{foot}-(\lambda,2,3))\cdot(1,2,1)=0$ gives $\lambda = 2$.
Two parallel lines, direction $\vec{d} = (2,3,6)$ ($|\vec{d}|=7$), through $(1,2,-4)$ and $(2,3,-5)$. Connector $\vec{w} = (1,1,-1)$:
$$\vec{w}\times\vec{d} = (1\cdot6-(-1)\cdot3,\ (-1)\cdot2-1\cdot6,\ 1\cdot3-1\cdot2) = (9,\ -8,\ 1)$$$$\text{dist} = \frac{|\vec{w}\times\vec{d}|}{|\vec{d}|} = \frac{\sqrt{81+64+1}}{7} = \frac{\sqrt{146}}{7}$$Answer: C ($\frac{\sqrt{146}}{7}$)
Solution
Directions: $\vec{d_2} = (1,2,2)$, $\vec{d_3} = (2,2,1)$.
$$\vec{d_1} = \vec{d_2}\times\vec{d_3} = (-2,\ 3,\ -2)$$$L_1 : r(-2,3,-2)$. Intersect with $L_2 : (3+t,\ 2t-1,\ 2t+4)$. Solving $r(-2,3,-2)=(3+t,2t-1,2t+4)$ gives $r=-1,\ t=-1$, so the intersection is $I = (2,\ -3,\ 2)$.
Point on $L_3$: $(3+2s,\ 3+2s,\ 2+s)$ at distance $\sqrt{17}$ from $I$:
$$(1+2s)^2 + (6+2s)^2 + s^2 = 17 \implies 9s^2 + 28s + 20 = 0 \implies s = -2\ \text{or}\ s = -\tfrac{10}{9}$$$s = -2$ gives $(a,b,c) = (-1,-1,0)$ with $a\in\mathbb{Z}$.
$$(a+b+c)^2 = (-2)^2 = 4$$Answer: 4