Three Dimensional Geometry JEE Main previous year questions with solutions

4 solved JEE Main questions on Three Dimensional Geometry, free to read — no sign-in needed. The full chapter has 209 questions; sign in to attempt the remaining 205 in the exam simulator.

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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 2020 (09 Jan, Shift 1)System of Linear Equations
   If for some $\alpha$ and $\beta$ in $R$ , the intersection of the following three planes $x+4y-2z=1$ $x+7y-5z=\beta$ $x+5y+\alpha z=5$ is a line in ${R}^{3}$ , then $\alpha +\beta$ is equal to:
   1. A.$0$
   2. B.$10$
   3. C.$2$
   4. D.$-10$
   Show answer & solution

   Answer: (B)

   $∆=0\Rightarrow \left|\begin{matrix}1 & 4 & -2 \\ 1 & 7 & -5 \\ 1 & 5 & \alpha \end{matrix}\right|=0$ $\left(7\alpha +25\right)-\left(4\alpha +10\right)+\left(-20+14\right)=0$ $3\alpha +9=0\Rightarrow \alpha =-3$ Also ${D}_{z}=0\Rightarrow \left|\begin{matrix}1 & 4 & 1 \\ 1 & 7 & \beta \\ 1 & 5 & 5\end{matrix}\right|=0$ $1\left(35-5\beta \right)-\left(15\right)+1\left(4\beta -7\right)=0$ $\beta =13$