Introduction to Three Dimensional Geometry JEE Main PYQs (2020–2026)

Solved Introduction to Three Dimensional Geometry PYQs

3 recent questions with answers. The full set of 297 is free to practise.

1. JEE Main 2026HardAngle between two lines
   Q1

   Let the direction cosines of two lines satisfy the equations : $4l+m-n=0$ and $2mn+10nl+3lm=0$ .
   
   Then the cosine of the acute angle between these lines is :

   • A$\frac{10}{7\sqrt{38}}$
   • B$\frac{10}{\sqrt{38}}$
   • C$\frac{10}{3\sqrt{38}}$
   • D$\frac{20}{3\sqrt{38}}$

   Show answer & solution

   Correct answer: (C)

   $$\begin{array}{rl}& n=4I+m\\ & n(2m+10I)+3Im=0\\ & \Rightarrow 2(4I+m)(m+5I)+3Im=0\\ & \Rightarrow 8Im+2m^2+10ml+40l^2+3Im=0\end{array}$$
   
   $$\begin{array}{rl}& \Rightarrow 40l^2+21ml+2m^2=0\\ & \Rightarrow 40l^2+16ml+5ml+2m^2=0\\ & \Rightarrow (8l+m)(5l+2m)=0\end{array}$$
   
   $$m=-8I;2m=-5I$$
   
   $$n=-4i\Rightarrow m=\frac{-5l}{2}$$
   
   $$\begin{array}{rl}& \begin{array}{rl}D\cdot R_1& =\left(l_1,m_1,n_1\right)\\ & =(-1,-8,-4)\\ D{R}_2& =\left(l_2,m_2,n_2\right)\\ & =\left(l_2,\frac{-5l_2}{2},\frac{3l_2}{2}\right)\\ =(2,-5,3)& \\ \cos \theta & =\left|\frac{2+40-12}{9\cdot \sqrt{38}}\right|\\ \cos \theta & =\frac{10}{3\sqrt{38}}\end{array}\\ & \text{ i.e. option (1) is correct }\end{array}$$
2. JEE Main 2026MediumEquation of a line in different forms (vector and Cartesian)
   Q2

   Let L be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let S be the set of all points $(a,b,c)$ on L , whose distance from the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$ along the line $L$ is 7 . Then ${\sum }_{(a,b,c)\in S}^{}(a+b+c)$ is equal to :

   • A28
   • B6
   • C40
   • D34

   Show answer & solution

   Correct answer: (D)

   M is the point of intersection of $L_1\&L_2$
   
   $$\begin{array}{rl}& \Rightarrow 2\lambda -1=2\mu -1,3\lambda -1=3\mu -1,6\lambda -3=9\Rightarrow \lambda =2=\mu \\ & M(3,5,9)\end{array}$$
   
   Now let point P be $(2K-1,3K-1,6K-3)$ on $L_2$
   
   Such that $PM=7$
   
   $$\begin{array}{rl}& \begin{array}{rl}& \Rightarrow \sqrt{(2K-4{)}^2+(3K-6)+(6K-12{)}^2}=7\\ & \Rightarrow 49K^2+196-196K=49\\ & \Rightarrow K^2+4-4K=1\Rightarrow K^2-4K+3=0\Rightarrow K=1,3\end{array}\\ & \end{array}$$
   
   $$\text{So points }P\&Q\text{ are }(1,2,3)\&(5,8,15)$$
   
   $$\text{So sum of all co-ordinates of }P\&Q=34$$
3. JEE Main 2026HardShortest distance between two skew lines
   Q3

   Let $P(\alpha ,\beta ,\gamma )$ be the point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4\sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance, between the lines $\frac{x-\alpha }{1}=\frac{y-\beta }{2}=\frac{z-\gamma }{3}$ and $\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$ , is equal to

   • A$4\sqrt{\frac{7}{5}}$
   • B$7\sqrt{\frac{5}{4}}$
   • C$4\sqrt{\frac{5}{7}}$
   • D$2\sqrt{\frac{7}{4}}$

   Show answer & solution

   Correct answer: (A)

   $$\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z}{1}=t$$
   
   $$\begin{array}{rl}& (1+2t,-1-3t,t)\\ & 4t^2+9t^2+t^2=16\times 14\\ & t=\pm 4\\ & \text{ Points are }(9,-13,4),(-7,11,-4)\\ & (\alpha ,\beta ,\gamma )=(-7,11,-4)\\ & \bar{a}=(-7,11,-4),\bar{c}=(-5,10,3)\\ & \bar{c}-\bar{a}=(2,-1,7)\\ & b\times d=(-1,5,-3)\\ & S.D=\left|\frac{-2-5-21}{\sqrt{1+25+9}}\right|=\frac{28}{\sqrt{35}}\end{array}$$

Topic-wise question count

Where JEE Main has asked from Introduction to Three Dimensional Geometry, most frequent first.

• Equation of a line in different forms (vector and Cartesian)108
• Equation of a plane in different forms (vector and Cartesian)60
• Shortest distance between two skew lines49
• Distance of a point from a plane48
• Angle between two lines13
• Direction cosines and direction ratios of a line10
• Angle between a line and a plane6
• Angle between two planes3

Year-wise Introduction to Three Dimensional Geometry questions in JEE Main

Number of questions from this chapter in each year's paper(s).