Three Dimensional Geometry JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Three Dimensional Geometry, free to read — no sign-in needed. The full chapter has 39 questions; sign in to attempt the remaining 34 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   Let $L$ be the straight line joining the points $P(1, 2, -1)$ and $Q(2, 3, 1)$. Let $S$ be the foot of the perpendicular drawn from the point $R(4, -1, 5)$ to the line $L$. Another line passing through $R$ intersects $L$ at a point $T$ such that the point $S$ divides the line segment $PT$ internally in the ratio $|PS| : |ST| = 1 : 2$, where $|PS|$ and $|ST|$ are the lengths of the line segments $PS$ and $ST$, respectively. Then which of the following statements is (are) TRUE ?
   1. A.The orthocentre of the triangle $PRT$ is $\left(\dfrac{23}{5}, -4, \dfrac{31}{5}\right)$
   2. B.The orthocentre of the triangle $PRT$ is $(4, 3, 5)$
   3. C.The area of the triangle $PRT$ is $6\sqrt{5}$
   4. D.The area of the triangle $PRT$ is $18\sqrt{5}$
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   Answer: A,D

   The direction ratios of line $L$ passing through $P(1, 2, -1)$ and $Q(2, 3, 1)$ are $(2-1, 3-2, 1-(-1)) = (1, 1, 2)$. The equation of line $L$ is $\dfrac{x-1}{1} = \dfrac{y-2}{1} = \dfrac{z+1}{2} = \lambda$. Any point on $L$ can be written as $S(\lambda+1, \lambda+2, 2\lambda-1)$. Since $S$ is the foot of the perpendicular from $R(4, -1, 5)$ to $L$, the direction ratios of $RS$ are $(\lambda-3, \lambda+3, 2\lambda-6)$. As $RS \perp L$, the dot product of their direction ratios is zero: $1(\lambda-3) + 1(\lambda+3) + 2(2\lambda-6) = 0$ $6\lambda - 12 = 0 \Rightarrow \lambda = 2$ Thus, the coordinates of $S$ are $(3, 4, 3)$. The length of $PS$ is $\sqrt{(3-1)^2 + (4-2)^2 + (3-(-1))^2} = \sqrt{4+4+16} = 2\sqrt{6}$. The length of the altitude $RS$ is $\sqrt{(3-4)^2 + (4-(-1))^2 + (3-5)^2} = \sqrt{1+25+4} = \sqrt{30}$. Since $S$ divides $PT$ internally in the ratio $1:2$, we have: $S = \dfrac{T + 2P}{3} \Rightarrow T = 3S - 2P$ $T = 3(3, 4, 3) - 2(1, 2, -1) = (7, 8, 11)$ The length of the base $PT$ is $3|PS| = 6\sqrt{6}$. The area of $\triangle PRT$ is $\dfrac{1}{2} \times |PT| \times |RS| = \dfrac{1}{2} \times 6\sqrt{6} \times \sqrt{30} = 18\sqrt{5}$. The orthocentre $H$ lies on the altitude $RS$. The line $RS$ passes through $S(3, 4, 3)$ and has direction ratios proportional to $R-S = (1, -5, 2)$. Let the coordinates of $H$ be $(3+t, 4-5t, 3+2t)$. Since $H$ is the orthocentre, $PH \perp RT$. The direction ratios of $RT$ are $(7-4, 8-(-1), 11-5) = (3, 9, 6)$, which is proportional to $(1, 3, 2)$. The direction ratios of $PH$ are $(3+t-1, 4-5t-2, 3+2t-(-1)) = (t+2, 2-5t, 2t+4)$. Taking the dot product of $PH$ and $RT$: $1(t+2) + 3(2-5t) + 2(2t+4) = 0$ $t + 2 + 6 - 15t + 4t + 8 = 0$ $16 - 10t = 0 \Rightarrow t = \dfrac{8}{5}$ Substituting $t = \dfrac{8}{5}$ into the coordinates of $H$, we get: $H = \left(3 + \dfrac{8}{5}, 4 - 5\left(\dfrac{8}{5}\right), 3 + 2\left(\dfrac{8}{5}\right)\right) = \left(\dfrac{23}{5}, -4, \dfrac{31}{5}\right)$ Answer: The orthocentre of the triangle $PRT$ is $\left(\dfrac{23}{5}, -4, \dfrac{31}{5}\right)$; The area of the triangle $PRT$ is $18\sqrt{5}$
2. Q2JEE Advanced Adv 2026 (Paper 1)
   Let $P$ be the plane such that it contains the straight line $\dfrac{x-1}{2} = \dfrac{y-3}{3} = \dfrac{z+2}{1}$ and is perpendicular to the plane $x + 2y + 3z = 4$. Let $P_1$ be the plane which passes through the point $(4, 2, 2)$ and is parallel to $P$. Then which of the following statements is (are) TRUE?
   1. A.The equation of the plane $P$ is $7x - 5y + z = -10$
   2. B.The distance between the planes $P$ and $P_1$ is $30$
   3. C.The distance of the plane $P$ from the origin is $2\sqrt{3}$
   4. D.The acute angle between the plane $P$ and the plane $2x + 2y + z = 3$ is $\cos^{-1}\left(\dfrac{1}{3\sqrt{3}}\right)$
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   Answer: A,D

   Let the normal vector to the plane $P$ be $\vec{n}$. Since the plane $P$ contains the line $\dfrac{x-1}{2} = \dfrac{y-3}{3} = \dfrac{z+2}{1}$, its normal vector $\vec{n}$ is perpendicular to the line's direction vector $\vec{b} = 2\hat{i} + 3\hat{j} + \hat{k}$. Since $P$ is perpendicular to the plane $x + 2y + 3z = 4$, $\vec{n}$ is perpendicular to its normal vector $\vec{n}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$. $\vec{n} = (2\hat{i} + 3\hat{j} + \hat{k}) \times (\hat{i} + 2\hat{j} + 3\hat{k}) = \hat{i}(9-2) - \hat{j}(6-1) + \hat{k}(4-3) = 7\hat{i} - 5\hat{j} + \hat{k}$ Plane $P$ passes through the point $(1, 3, -2)$ which lies on the given line. Equation of plane $P$ is $7(x-1) - 5(y-3) + 1(z+2) = 0 \Rightarrow 7x - 5y + z = -10$ Statement (1) is TRUE. The plane $P_1$ is parallel to $P$ and passes through $(4, 2, 2)$. Equation of plane $P_1$ is $7(x-4) - 5(y-2) + 1(z-2) = 0 \Rightarrow 7x - 5y + z = 20$ Distance between parallel planes $P$ and $P_1$ is $d = \dfrac{|20 - (-10)|}{\sqrt{7^2 + (-5)^2 + 1^2}} = \dfrac{30}{\sqrt{75}} = 2\sqrt{3}$. Statement (2) is FALSE. Distance of plane $P$ from the origin is $d_0 = \dfrac{|-10|}{\sqrt{75}} = \dfrac{2}{\sqrt{3}}$. Statement (3) is FALSE. Angle $\theta$ between the planes $7x - 5y + z + 10 = 0$ and $2x + 2y + z - 3 = 0$ is given by: $\cos\theta = \dfrac{|7(2) + (-5)(2) + 1(1)|}{\sqrt{7^2+(-5)^2+1^2} \sqrt{2^2+2^2+1^2}} = \dfrac{5}{\sqrt{75} \sqrt{9}} = \dfrac{5}{15\sqrt{3}} = \dfrac{1}{3\sqrt{3}}$ $\theta = \cos^{-1}\left(\dfrac{1}{3\sqrt{3}}\right)$. Statement (4) is TRUE.
3. Q3JEE Advanced Adv 2022 (Paper 1)
   Let $S$ be the reflection of a point $Q$ with respect to the plane given by $\vec{r}=-\left(t+p\right)\hat{i}+t\hat{j}+\left(1+p\right)\hat{k}$ where $t,p$ are real parameters and $\hat{i},\hat{j},\hat{k}$ are the unit vectors along the three positive coordinate axes. If the position vectors of $Q$ and $S$ are $10\hat{i}+15\hat{j}+20\hat{k}$ and $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$ respectively, then which of the following is/are TRUE?
   1. A.$3\left(\alpha +\beta \right)=-101$
   2. B.$3\left(\beta +\gamma \right)=-71$
   3. C.$3\left(\gamma +\alpha \right)=-86$
   4. D.$3\left(\alpha +\beta +\gamma \right)=-121$
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   Answer: A,B,C

   Given, Equation of plane is $\vec{r}=-\left(t+p\right)\hat{i}+t\hat{j}+\left(1+p\right)\hat{k}$ On rearranging we get, $\vec{r}=\hat{k}+t\left(-\hat{i}+\hat{j}\right)+p\left(-\hat{i}+\hat{k}\right)$ So, equation of plane in standard form is given by $\left[\begin{matrix}\vec{r}-\hat{k} & -\hat{i}+\hat{j} & -\hat{i}+\hat{k}\end{matrix}\right]=0$ $\Rightarrow \left|\begin{matrix}x & y & z-1 \\ -1 & 1 & 0 \\ -1 & 0 & 1\end{matrix}\right|=0$ $\Rightarrow x+y+z=1\cdots \left(1\right)$ Now given coordinate of $Q=\left(10,15,20\right)$ And coordinates of $S=\left(\alpha ,\beta ,\gamma \right)$ Now using the image formula of point and plane we get, $\dfrac{\alpha -10}{1}=\dfrac{\beta -15}{1}=\dfrac{\gamma -20}{1}=\dfrac{-2\left(10+15+20-1\right)}{3}$ $\Rightarrow \alpha -10=\beta -15=\gamma -20=-\dfrac{88}{3}$ $\Rightarrow \alpha =-\dfrac{58}{3},\beta =-\dfrac{43}{3},\gamma =-\dfrac{28}{3}$ Now solving all options we get, $3\left(\alpha +\beta \right)=-101$ $3\left(\beta +\gamma \right)=-71$ $3\left(\gamma +\alpha \right)=-86$ $3\left(\alpha +\beta +\gamma \right)=-129$
4. Q4JEE Advanced Adv 2020 (Paper 1)
   Let ${L}_{1}$ and ${L}_{2}$ be the following straight lines. ${L}_{1}:\dfrac{x-1}{1}=\dfrac{y}{-1}=\dfrac{z-1}{3}$ and ${L}_{2}:\dfrac{x-1}{-3}=\dfrac{y}{-1}=\dfrac{z-1}{1}$ Suppose the straight line $L:\dfrac{x-\alpha }{l}=\dfrac{y-1}{m}=\dfrac{z-\gamma }{-2}$ lies in the plane containing ${L}_{1}$ and ${L}_{2}$, and passes through the point of intersection of ${L}_{1}$ and ${L}_{2}$. If the line $L$ bisects the acute angle between the lines ${L}_{1}$ and ${L}_{2}$, then which of the following statements is/are TRUE?
   1. A.$\alpha -\gamma =3$
   2. B.$l+m=2$
   3. C.$\alpha -\gamma =1$
   4. D.$l+m=0$
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   Answer: A,B

   Given the lines ${L}_{1}:\dfrac{x-1}{1}=\dfrac{y}{-1}=\dfrac{z-1}{3}$ and ${L}_{2}:\dfrac{x-1}{-3}=\dfrac{y}{-1}=\dfrac{z-1}{1}$. From the given lines, we can say that both the lines pass through the point $\left(1,0,1\right)$. Now, let $\vec{a}$ be the direction vector of line ${L}_{1}$ i.e., $\vec{a}=i-j+3k$ and let $\vec{b}$ be the direction vector of line ${L}_{2}$ i.e., $\vec{b}=-3i-j+k$. We also know that the angle bisectors lie in the direction of $\left(\vec{a}+\vec{b}\right)\text{or}\left(\vec{a}-\vec{b}\right)$. Now, $\vec{a}.\vec{b}=\left(1\times -3\right)+\left(-1\times -1\right)+\left(3\times 1\right)=-3+1+3=1\gt 0$. So, direction ratios of the acute angle bisector between two lines will be in the direction of $\vec{a}+\vec{b}$ i.e., $\left(1-3,-1-1,3+1\right)=\left(-2,-2,4\right)=\left(1,1,-2\right)$. Since, $L:\dfrac{x-\alpha }{l}=\dfrac{y-1}{m}=\dfrac{z-\gamma }{-2}$ is the acute angle bisector between the lines ${L}_{1}\&{L}_{2}$. So, by comparing the direction ratios, we get $l=1\&m=1$ $\Rightarrow l+m=2$ So, equation of line $L$ will be $\dfrac{x-\alpha }{1}=\dfrac{y-1}{1}=\dfrac{z-\gamma }{-2}$. Given that the line $L$ passes through the point $\left(1,0,1\right)$. Putting the point in line $L$, we get $\dfrac{1-\alpha }{1}=\dfrac{0-1}{1}=\dfrac{1-\gamma }{-2}$ $\Rightarrow 1-\alpha =-1=\dfrac{1-\gamma }{-2}$ $\Rightarrow 1-\alpha =-1\&\dfrac{1-\gamma }{-2}=-1$ $\Rightarrow \alpha =2\&\gamma =-1$ $\Rightarrow \alpha -\gamma =3$ So, options $\left(A\right)\&\left(B\right)$ are correct.
5. Q5JEE Advanced Adv 2019 (Paper 2)
   Three lines ${L}_{1}:\vec{r}=\lambda \hat{i},\lambda \in R,$ ${L}_{2}:\vec{r}=\hat{k}+\mu \hat{j},\mu \in R$ and ${L}_{3}:\vec{r}=\hat{i}+\hat{j}+\nu \hat{k},\nu \in R$ are given. For which point(s) $Q$ and ${L}_{2}$ can we find a point $P$ on ${L}_{1}$ and a point $R$ on ${L}_{3}$ so that $P,Q$ and $R$ are collinear?
   1. A.$\hat{k}+\hat{j}$
   2. B.$\hat{k}$
   3. C.$\hat{k}+\dfrac{1}{2}\hat{j}$
   4. D.$\hat{k}-\dfrac{1}{2}\hat{j}$
   Show answer & solution

   Answer: C,D

   As given ${L}_{1}\Rightarrow \vec{r}=\lambda i,\lambda \in R$ ${L}_{2}\Rightarrow \vec{r}=k+\mu j,\mu \in R$ and ${L}_{3}\Rightarrow \vec{r}=i+j+\nu k,\nu \in R$ Let point $P\left(\lambda ,0,0\right),Q\left(0,\mu ,1\right)$ and $R\left(1,1,\nu \right)$ and $\vec{PQ}=-\lambda \hat{i}+\mu \hat{j}+\hat{k},\vec{PR}=\left(1-\lambda \right)\hat{i}+\hat{j}+\nu \hat{k}$ as $P,Q$ and $R$ are co-linear $\Rightarrow \vec{PQ}$ is parallel to $\vec{PR}$ $\Rightarrow -\dfrac{\lambda }{1-\lambda }=\dfrac{\mu }{1}=\dfrac{1}{\nu }$ $\Rightarrow \lambda =\dfrac{\mu }{\mu -1}$ and $\nu =\dfrac{1}{\mu }$ hence $\begin{matrix}\mu \neq 0 & \mu \neq 1\end{matrix}$ $Q\neq \left(0,0,1\right)$ and $Q\neq \left(0,1,1\right)$ $Q\neq \hat{k}\&Q\neq \left(\hat{j}+\hat{k}\right)$