Three Dimensional Geometry JEE Main previous year questions with solutions

5 solved JEE Main questions on Three Dimensional Geometry, free to read — no sign-in needed. The full chapter has 203 questions; sign in to attempt the remaining 198 in the exam simulator.

1. Q1JEE Main 2026 (23 Jan, Shift 1)Direction Cosines and Direction Ratios
   Let the direction cosines of two lines satisfy the equations : $4 l+m-n=0$ and $2 m n+10 n l+3 l m=0$. Then the cosine of the acute angle between these lines is :
   1. A.$\frac{20}{3 \sqrt{38}}$
   2. B.$\frac{10}{3 \sqrt{38}}$
   3. C.$\frac{10}{7 \sqrt{38}}$
   4. D.$\frac{10}{\sqrt{38}}$
   Show answer & solution

   Answer: (B)

   From $n = 4l + m$, substitute in $2mn + 10nl + 3lm = 0$: $40l^2 + 21lm + 2m^2 = 0$. Let $t = l/m$: $40t^2 + 21t + 2 = 0$. $t = \frac{-21 \pm 11}{80}$, giving $t = -1/8$ or $t = -2/5$. $l/m = -1/8$: direction $(l,m,n) = (-1, 8, 4)$. $l/m = -2/5$: direction $(l,m,n) = (-2, 5, -3)$. $\cos\theta = \frac{|2+40-12|}{\sqrt{81}\sqrt{38}} = \frac{30}{9\sqrt{38}} = \frac{10}{3\sqrt{38}}$.
2. Q2JEE Main 2025 (28 Jan, Shift 1)Basics of point in 3 dimension
   <p>Let $\mathrm{A}(x, y, z)$ be a point in $x y$-plane, which is equidistant from three points $(0,3,2),(2,0,3)$ and ( $0,0,1$ ). Let $\mathrm{B}=(1,4,-1)$ and $\mathrm{C}=(2,0,-2)$. Then among the statements (S1) : $\triangle \mathrm{ABC}$ is an isosceles right angled triangle, and (S2) : the area of $\triangle \mathrm{ABC}$ is $\frac{9 \sqrt{2}}{2}$,</p>
   1. A.<p>both are true</p>
   2. B.only (S2) is true
   3. C.only (S1) is true
   4. D.<p>both are false</p>
   Show answer & solution

   Answer: (C)

   <p>$\begin{aligned} & \mathrm{A}(\mathrm{x}, \mathrm{y}, \mathrm{z}) \text { Let } \mathrm{P}(0,3,2), \mathrm{Q}(2,0,3), \mathrm{R}(0,0,1) \\ & \mathrm{AP}=\mathrm{AQ}=\mathrm{AR} \\ & \mathrm{x}^2+(\mathrm{y}-3)^2+(\mathrm{z}-2)^2=(\mathrm{x}-2)^2+\mathrm{y}^2+(\mathrm{z}-3)^2=\mathrm{x}^2+ \\ & \mathrm{y}^2+(\mathrm{z}-1)^2 \end{aligned}$ In $x y$ plane $z=0$ So, $x^2-4 x+4+y^2+9=x^2+y^2+1$ $\begin{aligned} & x=3 \\ & 9+y^2-6 y+9+4=x^2+y^2+1 \end{aligned}$ So, $\mathrm{A}(3,2,0)$ also $\mathrm{B}(1,4,-1) \& \mathrm{C}(2,0,-2)$ Now $A B=\sqrt{4+4+1}=3$ $\begin{aligned} & \mathrm{AC}=\sqrt{1+4+4}=3 \\ & \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18} \end{aligned}$ $\mathrm{AB}=\mathrm{AC}$ isosceles $\Delta \& \mathrm{AB}^2+\mathrm{AC}^2=\mathrm{BC}^2$ right angle $\Delta$ Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times$ base.height $\frac{1}{2} \times 3 \times 3=\frac{9}{2}$ So only $S_1$ is true</p>
3. Q3JEE Main 2024 (08 Apr, Shift 2)Line in Space
   If the shortest distance between the lines $\frac{x-\lambda}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$, then a value of $\lambda$ is :
   1. A.-1
   2. B.$-\frac{13}{25}$
   3. C.$\frac{13}{25}$
   4. D.1
   Show answer & solution

   Answer: (D)

   $\left.\begin{array}{l}\overline{\mathrm{r}}_1=(\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\alpha(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \\ \overline{\mathrm{r}}_2=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})+\beta(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})\end{array}\right\} \begin{gathered}\overline{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ \overline{\mathrm{a}}_2+\lambda \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ \mathrm{a}_2=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}\end{gathered}$ Shortest dist. $=\frac{\left|\overline{\mathrm{b}} \times\left(\overline{\mathrm{a}}_2-\overline{\mathrm{a}}_1\right)\right|}{|\mathrm{b}|}=\frac{13}{\sqrt{29}}$ $\begin{aligned} & \left\lvert\, \frac{|(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times((2-\lambda) \hat{i}+4 \hat{k})|}{\sqrt{29}}=\frac{13}{\sqrt{29}}\right. \\ & |-8 \hat{\mathrm{j}}-3(2-\lambda) \hat{k}+12 \hat{i}+4(2-\lambda) \hat{j}|=13 \\ & |12 \hat{i}-4 \lambda \hat{j}+(3 \lambda-6) \hat{k}|=13\end{aligned}$ $\begin{aligned} & 144+16 \lambda^2+(3 \lambda-6)^2=169 \\ & 16 \lambda^2+(3 \lambda-6)^2=25=\lambda \Rightarrow=1\end{aligned}$
4. Q4JEE Main 2026 (08 Apr, Shift 2)Line in Space
   Let a line $L_1$ pass through the origin and be perpendicular to the lines $L_2: \vec{r} = (3+t)\hat{i} + (2t-1)\hat{j} + (2t+4)\hat{k}$ and $L_3: \vec{r} = (3+2s)\hat{i} + (3+2s)\hat{j} + (2+s)\hat{k}$, $t, s \in \mathbb{R}$. If $(a, b, c)$, $a \in \mathbb{Z}$, is the point on $L_3$ at a distance of $\sqrt{17}$ from the point of intersection of $L_1$ and $L_2$, then $(a+b+c)^2$ is equal to ________.
   Show answer & solution

   Answer: 4

   The direction vectors of the given lines $L_2$ and $L_3$ are $\vec{d_2} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{d_3} = 2\hat{i} + 2\hat{j} + \hat{k}$ respectively. Since line $L_1$ is perpendicular to both $L_2$ and $L_3$, its direction vector $\vec{d_1}$ is given by the cross product of $\vec{d_2}$ and $\vec{d_3}$: $\vec{d_1} = \vec{d_2} \times \vec{d_3} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix} = -2\hat{i} + 3\hat{j} - 2\hat{k}$ Since $L_1$ passes through the origin, its equation is: $\vec{r} = \lambda(-2\hat{i} + 3\hat{j} - 2\hat{k})$ Let the point of intersection of $L_1$ and $L_2$ be $P$. The coordinates of $P$ can be written as $(-2\lambda, 3\lambda, -2\lambda)$ from $L_1$ and $(3+t, 2t-1, 2t+4)$ from $L_2$. Equating the respective coordinates: $-2\lambda = 3 + t$ $3\lambda = 2t - 1$ $-2\lambda = 2t + 4$ From the first and third equations, we get: $3 + t = 2t + 4 \Rightarrow t = -1$ Substituting $t = -1$ into the first equation gives $-2\lambda = 2 \Rightarrow \lambda = -1$. This also satisfies the second equation. Thus, the point of intersection is $P(2, -3, 2)$. Let the point on $L_3$ be $Q(a, b, c) = (3+2s, 3+2s, 2+s)$. The distance between $P$ and $Q$ is given as $\sqrt{17}$. $PQ^2 = 17$ $(3+2s - 2)^2 + (3+2s - (-3))^2 + (2+s - 2)^2 = 17$ $(2s + 1)^2 + (2s + 6)^2 + s^2 = 17$ $4s^2 + 4s + 1 + 4s^2 + 24s + 36 + s^2 = 17$ $9s^2 + 28s + 37 = 17$ $9s^2 + 28s + 20 = 0$ $9s^2 + 18s + 10s + 20 = 0$ $(9s + 10)(s + 2) = 0$ This gives $s = -2$ or $s = -\dfrac{10}{9}$. Since $a = 3 + 2s$ must be an integer ($a \in \mathbb{Z}$), we must choose $s = -2$. Substituting $s = -2$ into the coordinates of $Q$: $a = 3 + 2(-2) = -1$ $b = 3 + 2(-2) = -1$ $c = 2 + (-2) = 0$ Therefore, $(a, b, c) = (-1, -1, 0)$. We need to find the value of $(a+b+c)^2$: $(a+b+c)^2 = (-1 - 1 + 0)^2 = (-2)^2 = 4$ Answer: $4$
5. Q5JEE Main 2026 (08 Apr, Shift 2)Line in Space
   Let the foot of perpendicular from the point $(\lambda, 2, 3)$ on the line $\dfrac{x-4}{1} = \dfrac{y-9}{2} = \dfrac{z-5}{1}$ be the point $(1, \mu, 2)$. Then the distance between the lines $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z+4}{6}$ and $\dfrac{x-\lambda}{2} = \dfrac{y-\mu}{3} = \dfrac{z+5}{6}$ is equal to:
   1. A.$\dfrac{12}{7}$
   2. B.$\dfrac{\sqrt{145}}{7}$
   3. C.$\dfrac{\sqrt{146}}{7}$
   4. D.$\dfrac{\sqrt{143}}{7}$
   Show answer & solution

   Answer: (C)

   Since the point $(1, \mu, 2)$ lies on the line $\dfrac{x-4}{1} = \dfrac{y-9}{2} = \dfrac{z-5}{1}$, we substitute its coordinates into the line equation: $\dfrac{1-4}{1} = \dfrac{\mu-9}{2} = \dfrac{2-5}{1}$ $-3 = \dfrac{\mu-9}{2} = -3 \Rightarrow \mu - 9 = -6 \Rightarrow \mu = 3$ The direction ratios of the given line are $(1, 2, 1)$. The direction ratios of the perpendicular from $(\lambda, 2, 3)$ to $(1, 3, 2)$ are $(1-\lambda, 3-2, 2-3) = (1-\lambda, 1, -1)$. Since the lines are perpendicular, their dot product is zero: $1(1-\lambda) + 2(1) + 1(-1) = 0$ $1 - \lambda + 2 - 1 = 0 \Rightarrow \lambda = 2$ Substituting $\lambda = 2$ and $\mu = 3$ into the equations of the two lines, we get: $L_1: \dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z+4}{6}$ $L_2: \dfrac{x-2}{2} = \dfrac{y-3}{3} = \dfrac{z+5}{6}$ These are parallel lines with direction vector $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$. The lines pass through the points $\vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k}$ and $\vec{a}_2 = 2\hat{i} + 3\hat{j} - 5\hat{k}$ respectively. The distance $d$ between two parallel lines is given by $d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$. We have $\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (3-2)\hat{j} + (-5 - (-4))\hat{k} = \hat{i} + \hat{j} - \hat{k}$. Now, we compute the cross product: $(\vec{a}_2 - \vec{a}_1) \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 2 & 3 & 6 \end{vmatrix}$ $= \hat{i}(6 - (-3)) - \hat{j}(6 - (-2)) + \hat{k}(3 - 2) = 9\hat{i} - 8\hat{j} + \hat{k}$ The magnitude of this vector is: $|9\hat{i} - 8\hat{j} + \hat{k}| = \sqrt{9^2 + (-8)^2 + 1^2} = \sqrt{81 + 64 + 1} = \sqrt{146}$ The magnitude of the direction vector $\vec{b}$ is: $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ Therefore, the distance between the lines is: $d = \dfrac{\sqrt{146}}{7}$ Answer: $\dfrac{\sqrt{146}}{7}$