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3D Geometry Formulas For JEE 2027
Three-Dimensional Geometry in JEE Mathematics focuses on representing points and lines in space using coordinates and vector equations. Key concepts include direction ratios, direction cosines, equations of lines, angles between lines, and the shortest distance between skew lines. Understanding these ideas helps students translate spatial relationships into equations and solve problems systematically.
Start by learning how to identify a line’s direction and choose the appropriate equation or distance formula. Revise essential JEE Mains Formulas, then practise questions involving intersecting, parallel, and skew lines. Pay attention to the difference between direction ratios and direction cosines, and check calculations carefully. Taking a Free JEE Mains mock test can help you assess your understanding, identify recurring errors, and improve your speed and accuracy under exam conditions.
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 2027: Conclusion
A strong grasp of 3D Geometry comes from understanding direction ratios, direction cosines, equations of lines, and distance calculations. Practise identifying the relationship between lines before choosing a formula, and check your signs and vector components carefully. Keep a JEE Maths Formula sheet handy to revise key results and the conditions under which they apply.
When planning revision around JEE Mains Maths Important Chapters, give 3D Geometry regular practice time. Solve previous-year questions, revisit mistakes, and attempt timed problems to strengthen your understanding and improve calculation speed. Combining formula revision with consistent problem-solving can help you approach this chapter with greater accuracy and confidence.
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