3D Geometry Formulas For JEE 2026
Three-Dimensional Geometry is an important chapter in JEE Mains Mathematics and usually carries 2–3 questions every year. In this topic, students learn how to work with the 3D coordinate system and represent points, lines, and planes in space. Concepts like direction ratios and direction cosines help in understanding the direction and position of lines in three dimensions.
The chapter also covers equations of lines and planes, angles between them, and the shortest distance between skew lines, which are commonly asked in exams. Since many questions are formula-based, regular practice and revision are very important. For quick revision before exams, students can also use a well-organized JEE Mains Maths Formula PDF to go through important formulas and concepts easily.
3D Coordinate System
Distance Between Two Points
The distance between $$P(x_1, y_1, z_1)$$ and $$Q(x_2, y_2, z_2)$$:
$$PQ = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$Distance from origin: $$OP = \sqrt{x^2 + y^2 + z^2}$$
Worked Example
Find the distance between $$A(1, 2, 3)$$ and $$B(4, 6, 3)$$.
$$AB = \sqrt{(4-1)^2 + (6-2)^2 + (3-3)^2} = \sqrt{9 + 16 + 0} = \sqrt{25} = \textbf{5}$$
Section Formula in 3D
Section Formula
If $$P$$ divides the join of $$A(x_1,y_1,z_1)$$ and $$B(x_2,y_2,z_2)$$:
Internally in the ratio $$m:n$$:
$$P = \left(\frac{mx_2+nx_1}{m+n},\; \frac{my_2+ny_1}{m+n},\; \frac{mz_2+nz_1}{m+n}\right)$$Externally in the ratio $$m:n$$:
$$P = \left(\frac{mx_2-nx_1}{m-n},\; \frac{my_2-ny_1}{m-n},\; \frac{mz_2-nz_1}{m-n}\right)$$Midpoint:
$$M = \left(\frac{x_1+x_2}{2},\; \frac{y_1+y_2}{2},\; \frac{z_1+z_2}{2}\right)$$Direction Cosines and Direction Ratios
Direction Cosines
If a directed line makes angles $$\alpha$$, $$\beta$$, $$\gamma$$ with the positive $$x$$, $$y$$, $$z$$-axes respectively, then $$\cos\alpha$$, $$\cos\beta$$, $$\cos\gamma$$ are its direction cosines (DCs), denoted $$l$$, $$m$$, $$n$$:
$$l = \cos\alpha, \quad m = \cos\beta, \quad n = \cos\gamma$$
Fundamental Relation of Direction Cosines
$$l^2 + m^2 + n^2 = 1$$Direction cosines are like a unit vector specifying direction.
Direction Ratios
Any three numbers $$a$$, $$b$$, $$c$$ proportional to the direction cosines are called direction ratios (DRs). If $$a$$, $$b$$, $$c$$ are DRs, then:
$$l = \dfrac{a}{\sqrt{a^2+b^2+c^2}}, \quad m = \dfrac{b}{\sqrt{a^2+b^2+c^2}}, \quad n = \dfrac{c}{\sqrt{a^2+b^2+c^2}}$$
DRs of the Line Joining Two Points
The direction ratios of the line joining $$A(x_1,y_1,z_1)$$ and $$B(x_2,y_2,z_2)$$:
$$a = x_2 - x_1, \quad b = y_2 - y_1, \quad c = z_2 - z_1$$Worked Example
Find the direction cosines of the line joining $$A(1, 2, 3)$$ and $$B(3, 5, 7)$$.
Direction ratios: $$a = 2$$, $$b = 3$$, $$c = 4$$. $$\sqrt{a^2+b^2+c^2} = \sqrt{29}$$
Direction cosines: $$l = \dfrac{2}{\sqrt{29}},\; m = \dfrac{3}{\sqrt{29}},\; n = \dfrac{4}{\sqrt{29}}$$
Tip: If a question gives direction ratios $$(a, b, c)$$ and asks for direction cosines, just divide each by $$\sqrt{a^2+b^2+c^2}$$. This is exactly the same as finding a unit vector.
Equation of a Line in 3D
Vector Equation of a Line
A line passing through a point with position vector $$\vec{a}$$ and parallel to vector $$\vec{b}$$:
$$\vec{r} = \vec{a} + \lambda\vec{b}$$where $$\lambda$$ is a real parameter.
Cartesian Equation of a Line
A line through $$(x_1, y_1, z_1)$$ with direction ratios $$a$$, $$b$$, $$c$$:
$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$Line Through Two Points
The line through $$A(x_1,y_1,z_1)$$ and $$B(x_2,y_2,z_2)$$:
$$\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$$Vector form: $$\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$$
Worked Example
Find the Cartesian equation of the line through $$(2, -1, 3)$$ with direction ratios $$3, 2, -5$$.
$$\dfrac{x - 2}{3} = \dfrac{y + 1}{2} = \dfrac{z - 3}{-5}$$
Important: When one of the direction ratios is 0, say $$b = 0$$, write the equation as $$\dfrac{x-x_1}{a} = \dfrac{z-z_1}{c}$$, $$y = y_1$$. The line lies in the plane $$y = y_1$$.
Angle Between Two Lines
If the lines have direction ratios $$(a_1, b_1, c_1)$$ and $$(a_2, b_2, c_2)$$:
$$\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\;\sqrt{a_2^2+b_2^2+c_2^2}}$$Perpendicularity ($$\theta = 90°$$): $$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$$
Parallelism ($$\theta = 0°$$): $$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$$
Worked Example
Find the angle between the lines $$\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z}{1}$$ and $$\dfrac{x}{1} = \dfrac{y}{-1} = \dfrac{z+2}{2}$$.
Direction ratios: $$(2, 3, 1)$$ and $$(1, -1, 2)$$.
$$\cos\theta = \dfrac{|2(1) + 3(-1) + 1(2)|}{\sqrt{14}\;\sqrt{6}} = \dfrac{1}{\sqrt{84}} = \dfrac{1}{2\sqrt{21}}$$
$$\theta = \cos^{-1}\!\left(\dfrac{1}{2\sqrt{21}}\right)$$
Equation of a Plane
General Equation
$$ax + by + cz + d = 0$$where $$(a, b, c)$$ are direction ratios of the normal to the plane.
Normal Form (Vector)
$$\vec{r} \cdot \hat{n} = p$$where $$\hat{n}$$ is the unit normal and $$p$$ is the perpendicular distance from the origin.
Point-Normal Form
A plane through point $$A(x_1,y_1,z_1)$$ with normal direction ratios $$(a, b, c)$$:
$$a(x-x_1) + b(y-y_1) + c(z-z_1) = 0$$Vector form: $$(\vec{r} - \vec{a}) \cdot \vec{n} = 0$$
Intercept Form
If a plane makes intercepts $$a$$, $$b$$, $$c$$ on the $$x$$, $$y$$, $$z$$-axes:
$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$Three-Point Form
The plane through three non-collinear points $$A(x_1,y_1,z_1)$$, $$B(x_2,y_2,z_2)$$, $$C(x_3,y_3,z_3)$$:
$$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$$Worked Example
Find the equation of the plane passing through $$(1, 2, 3)$$ with normal vector $$2\hat{i} - \hat{j} + \hat{k}$$.
$$2(x-1) + (-1)(y-2) + 1(z-3) = 0 \Rightarrow \boldsymbol{2x - y + z = 3}$$
Worked Example
Find the equation of the plane through $$A(1,1,1)$$, $$B(1,2,3)$$, $$C(2,1,3)$$.
$$\begin{vmatrix} x-1 & y-1 & z-1 \\ 0 & 1 & 2 \\ 1 & 0 & 2 \end{vmatrix} = 0$$
$$2(x-1) + 2(y-1) - (z-1) = 0 \Rightarrow \boldsymbol{2x + 2y - z = 3}$$
Angle Between Two Planes
For planes $$a_1x + b_1y + c_1z + d_1 = 0$$ and $$a_2x + b_2y + c_2z + d_2 = 0$$:
$$\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\;\sqrt{a_2^2+b_2^2+c_2^2}}$$Perpendicular planes: $$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$$
Parallel planes: $$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$$
Distance of a Point from a Plane
The perpendicular distance from $$P(x_1, y_1, z_1)$$ to the plane $$ax + by + cz + d = 0$$:
$$D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$Distance Between Parallel Planes
Distance between $$ax+by+cz+d_1 = 0$$ and $$ax+by+cz+d_2 = 0$$:
$$D = \frac{|d_1 - d_2|}{\sqrt{a^2+b^2+c^2}}$$Worked Example
Find the distance of $$(2, 3, -5)$$ from the plane $$x + 2y - 2z = 9$$.
$$D = \dfrac{|1(2) + 2(3) + (-2)(-5) - 9|}{\sqrt{1+4+4}} = \dfrac{|2 + 6 + 10 - 9|}{3} = \dfrac{9}{3} = \textbf{3}$$
Line and Plane Interactions
Angle Between a Line and a Plane
If a line has direction ratios $$(a_1, b_1, c_1)$$ and the plane has normal $$(a_2, b_2, c_2)$$:
$$\sin\phi = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\;\sqrt{a_2^2+b_2^2+c_2^2}}$$Note: We use $$\sin\phi$$ (not $$\cos\theta$$) because the angle is measured from the plane, not from the normal.
Line parallel to plane: $$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$$
Line perpendicular to plane: $$\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$$
Coplanarity of Two Lines
Condition for Coplanarity
Lines $$\dfrac{x-x_1}{a_1} = \dfrac{y-y_1}{b_1} = \dfrac{z-z_1}{c_1}$$ and $$\dfrac{x-x_2}{a_2} = \dfrac{y-y_2}{b_2} = \dfrac{z-z_2}{c_2}$$ are coplanar if:
$$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$$Shortest Distance Between Skew Lines
Skew Lines
Two lines that do not lie in any common plane. They are not parallel and do not intersect.
Shortest Distance Between Skew Lines
For lines $$\vec{r} = \vec{a_1} + \lambda\vec{b_1}$$ and $$\vec{r} = \vec{a_2} + \mu\vec{b_2}$$:
$$d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$$In Cartesian form:
$$d = \frac{\left|\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}\right|}{\sqrt{(b_1 c_2 - b_2 c_1)^2 + (c_1 a_2 - c_2 a_1)^2 + (a_1 b_2 - a_2 b_1)^2}}$$If $$d = 0$$, the lines are coplanar.
Distance Between Parallel Lines
For parallel lines $$\vec{r} = \vec{a_1} + \lambda\vec{b}$$ and $$\vec{r} = \vec{a_2} + \mu\vec{b}$$:
$$d = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|}$$Worked Example
Find the shortest distance between the lines:
$$\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{-1}$$ and $$\dfrac{x+1}{5} = \dfrac{y-2}{1} = \dfrac{z}{3}$$
Points: $$(1, -1, 1)$$ and $$(-1, 2, 0)$$. DRs: $$(2, 3, -1)$$ and $$(5, 1, 3)$$.
$$\vec{b_1} \times \vec{b_2} = 10\hat{i} - 11\hat{j} - 13\hat{k}$$, $$|\vec{b_1} \times \vec{b_2}| = \sqrt{390}$$
$$(\vec{a_2}-\vec{a_1}) \cdot (\vec{b_1}\times\vec{b_2}) = -20 - 33 + 13 = -40$$
$$d = \dfrac{40}{\sqrt{390}}$$ units
Tip: For the shortest distance formula, the numerator is the scalar triple product of $$(\vec{a_2}-\vec{a_1})$$, $$\vec{b_1}$$, $$\vec{b_2}$$, and the denominator is $$|\vec{b_1} \times \vec{b_2}|$$. If the numerator is 0, the lines are coplanar.
Quick Reference
Summary Table
| Formula | Expression |
|---|---|
| Distance (3D) | $$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$ |
| DC relation | $$l^2+m^2+n^2 = 1$$ |
| Line (Cartesian) | $$\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$$ |
| Line (vector) | $$\vec{r} = \vec{a} + \lambda\vec{b}$$ |
| Plane (general) | $$ax+by+cz+d=0$$ |
| Plane (intercept) | $$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$$ |
| Point-to-plane dist. | $$\dfrac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}$$ |
| Line-plane angle | $$\sin\phi = \dfrac{|\vec{b}\cdot\vec{n}|}{|\vec{b}||\vec{n}|}$$ |
| Skew line dist. | $$\dfrac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|}$$ |
3D Geometry Formulas For JEE 2026: Conclusion
3D Geometry is one of the most important and scoring topics in JEE Mathematics, especially because many questions are directly based on formulas. Concepts like equations of lines and planes, direction ratios, and distances play a key role in solving problems quickly. With regular practice, students can easily master these concepts and improve their accuracy in the exam.
To perform well in this chapter, it is essential to focus on understanding the logic behind formulas rather than just memorizing them. Consistent revision and solving previous year questions will help in building confidence and speed. A well-structured formula sheet can be very useful for quick revision before the exam and ensures better retention of important concepts.
