Three Dimensional Geometry Formula Sheet
Every key 3D Geometry formula for JEE - distance, section, direction cosines, line & plane equations, angles, shortest distance & coplanarity. Quick revision.
All the must-know formulas from the Three Dimensional Geometry chapter in one scannable sheet - coordinates, direction cosines, lines, planes, angles, shortest distance and coplanarity. Use it for last-minute revision before JEE Main and Advanced.
Coordinates, Distance and Section Formula
Distance between $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$:
$$\boxed{AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}}$$Distance of $P(x, y, z)$ from the origin:
$$\boxed{OP = \sqrt{x^2 + y^2 + z^2}}$$Point dividing $AB$ internally in ratio $m:n$:
$$\boxed{P = \left(\frac{mx_2 + nx_1}{m+n},\ \frac{my_2 + ny_1}{m+n},\ \frac{mz_2 + nz_1}{m+n}\right)}$$Point dividing $AB$ externally in ratio $m:n$:
$$\boxed{P = \left(\frac{mx_2 - nx_1}{m-n},\ \frac{my_2 - ny_1}{m-n},\ \frac{mz_2 - nz_1}{m-n}\right)}$$Midpoint (special case $m = n = 1$):
$$\boxed{M = \left(\frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2},\ \frac{z_1 + z_2}{2}\right)}$$Centroid of a triangle $A, B, C$:
$$\boxed{G = \left(\frac{x_1 + x_2 + x_3}{3},\ \frac{y_1 + y_2 + y_3}{3},\ \frac{z_1 + z_2 + z_3}{3}\right)}$$Centroid of a tetrahedron $A, B, C, D$:
$$\boxed{G = \left(\frac{x_1 + x_2 + x_3 + x_4}{4},\ \frac{y_1 + y_2 + y_3 + y_4}{4},\ \frac{z_1 + z_2 + z_3 + z_4}{4}\right)}$$| Quantity | Formula | Notes |
|---|---|---|
| Distance $AB$ | $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ | Order of subtraction irrelevant |
| Distance from origin | $\sqrt{x^2 + y^2 + z^2}$ | Special case with $O$ |
| Internal division $m:n$ | $\left(\tfrac{mx_2+nx_1}{m+n}, \dots\right)$ | $+$ in numerator, $m+n$ below |
| External division $m:n$ | $\left(\tfrac{mx_2-nx_1}{m-n}, \dots\right)$ | $-$ in numerator, $m-n$ below |
| Midpoint | $\left(\tfrac{x_1+x_2}{2}, \dots\right)$ | Average of coordinates |
| Centroid (triangle) | $\left(\tfrac{x_1+x_2+x_3}{3}, \dots\right)$ | Divide by 3 |
| Centroid (tetrahedron) | $\left(\tfrac{x_1+x_2+x_3+x_4}{4}, \dots\right)$ | Divide by 4 |
Direction Cosines and Direction Ratios
For a line making angles $\alpha, \beta, \gamma$ with the positive X, Y, Z axes, the direction cosines are:
$$\boxed{l = \cos\alpha, \quad m = \cos\beta, \quad n = \cos\gamma}$$Fundamental identity (always holds):
$$\boxed{l^2 + m^2 + n^2 = 1}$$Equivalently $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$, which also gives $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$.
Direction cosines of the line joining $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$:
$$\boxed{l = \frac{x_2 - x_1}{AB}, \quad m = \frac{y_2 - y_1}{AB}, \quad n = \frac{z_2 - z_1}{AB}}$$where $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
Direction ratios $(a, b, c)$ are any triple proportional to $(l, m, n)$:
$$\boxed{a : b : c = l : m : n}$$Converting direction ratios to direction cosines:
$$\boxed{l = \frac{\pm a}{\sqrt{a^2+b^2+c^2}}, \quad m = \frac{\pm b}{\sqrt{a^2+b^2+c^2}}, \quad n = \frac{\pm c}{\sqrt{a^2+b^2+c^2}}}$$The $\pm$ accounts for the two opposite directions along the same line.
After computing direction cosines, always verify $l^2 + m^2 + n^2 = 1$. It instantly catches arithmetic slips.
| Object | Direction Cosines $(l, m, n)$ | Direction Ratios |
|---|---|---|
| X-axis | $(1, 0, 0)$ | $(1, 0, 0)$ |
| Y-axis | $(0, 1, 0)$ | $(0, 1, 0)$ |
| Z-axis | $(0, 0, 1)$ | $(0, 0, 1)$ |
Angle Between Two Lines
Using direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$:
$$\boxed{\cos\theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|}$$Using direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$:
$$\boxed{\cos\theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}}$$Using vector direction vectors $\vec{b_1}, \vec{b_2}$:
$$\boxed{\cos\theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}|\,|\vec{b_2}|}}$$The absolute value forces the acute angle $0^\circ \le \theta \le 90^\circ$.
| Condition | Direction Ratios | Direction Cosines |
|---|---|---|
| Parallel ($\theta = 0^\circ$) | $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$ | $l_1 = l_2,\ m_1 = m_2,\ n_1 = n_2$ |
| Perpendicular ($\theta = 90^\circ$) | $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$ | $l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$ |
Equation of a Line in Space
Vector form (through point with position vector $\vec{a}$, parallel to $\vec{b}$):
$$\boxed{\vec{r} = \vec{a} + \lambda\vec{b}}$$Two-point vector form (through $A$ and $B$):
$$\boxed{\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}) = (1-\lambda)\vec{a} + \lambda\vec{b}}$$Parametric form (point $(x_1, y_1, z_1)$, direction ratios $(a, b, c)$):
$$\boxed{x = x_1 + \lambda a, \quad y = y_1 + \lambda b, \quad z = z_1 + \lambda c}$$Symmetric / Cartesian form (most used in JEE):
$$\boxed{\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}}$$Two-point symmetric form:
$$\boxed{\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}}$$| Form | Equation | Best for |
|---|---|---|
| Vector | $\vec{r} = \vec{a} + \lambda\vec{b}$ | Theory, vector problems |
| Parametric | $x = x_1 + \lambda a,\ y = y_1 + \lambda b,\ z = z_1 + \lambda c$ | Finding specific points |
| Symmetric | $\dfrac{x-x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{z-z_1}{c}$ | Most calculations, point checks |
Equation of a Plane
General form - $(a, b, c)$ are the direction ratios of the normal:
$$\boxed{ax + by + cz + d = 0}$$Point-normal form (through $(x_1,y_1,z_1)$ with normal $(a,b,c)$):
$$\boxed{a(x - x_1) + b(y - y_1) + c(z - z_1) = 0}$$Vector form of point-normal:
$$\boxed{(\vec{r} - \vec{a}) \cdot \vec{n} = 0}$$Intercept form ($a, b, c$ are the X, Y, Z intercepts):
$$\boxed{\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1}$$Plane through three points $A, B, C$:
$$\boxed{\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ x_3 - x_1 & y_3 - y_1 & z_3 - z_1 \end{vmatrix} = 0}$$Parametric (vector) form ($\vec{b}, \vec{c}$ non-parallel and parallel to plane):
$$\boxed{\vec{r} = \vec{a} + \lambda\vec{b} + \mu\vec{c}}$$Plane through the origin: $\boxed{ax + by + cz = 0}$ (constant term $d = 0$).
Intercepts of $ax + by + cz + d = 0$: x-intercept $= -d/a$, y-intercept $= -d/b$, z-intercept $= -d/c$.
| Special plane | Equation | Normal |
|---|---|---|
| Parallel to XY-plane | $z = k$ | $(0, 0, 1)$ |
| Parallel to YZ-plane | $x = k$ | $(1, 0, 0)$ |
| Parallel to XZ-plane | $y = k$ | $(0, 1, 0)$ |
Line and Plane Relations
Angle between a line (direction ratios $l, m, n$) and a plane (normal $a, b, c$) - note the sine:
$$\boxed{\sin\theta = \frac{|al + bm + cn|}{\sqrt{a^2+b^2+c^2}\,\sqrt{l^2+m^2+n^2}}}$$| Relation | Condition |
|---|---|
| Line parallel to plane ($\theta = 0^\circ$) | $al + bm + cn = 0$ |
| Line perpendicular to plane ($\theta = 90^\circ$) | $\dfrac{l}{a} = \dfrac{m}{b} = \dfrac{n}{c}$ |
Distance from point $P(x_1, y_1, z_1)$ to plane $ax + by + cz + d = 0$:
$$\boxed{D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}}$$Distance between parallel planes $ax + by + cz + d_1 = 0$ and $ax + by + cz + d_2 = 0$:
$$\boxed{D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}}}$$(Coefficients of $x, y, z$ must be identical - make the normals match before applying.)
Image of point $(x_1, y_1, z_1)$ in plane $ax + by + cz + d = 0$, writing $k = ax_1 + by_1 + cz_1 + d$:
$$\boxed{\left(x_1 - \frac{2ak}{a^2+b^2+c^2},\ y_1 - \frac{2bk}{a^2+b^2+c^2},\ z_1 - \frac{2ck}{a^2+b^2+c^2}\right)}$$Intersection of line and plane - substitute the parametric line into the plane and solve for $\lambda$:
| Result for $\lambda$ | Geometric meaning |
|---|---|
| Unique value | Line meets plane at one point |
| $0 = 0$ (identity) | Line lies in the plane |
| $0 = k,\ k \ne 0$ | Line parallel to plane, no intersection |
Shortest Distance Between Lines
Skew lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ (vector form):
$$\boxed{SD = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}}$$Cartesian form:
$$\boxed{SD = \frac{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}}{\sqrt{(b_1 c_2 - b_2 c_1)^2 + (c_1 a_2 - c_2 a_1)^2 + (a_1 b_2 - a_2 b_1)^2}}}$$Parallel lines $\vec{r} = \vec{a_1} + \lambda\vec{b}$ and $\vec{r} = \vec{a_2} + \mu\vec{b}$:
$$\boxed{SD = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{b}|}{|\vec{b}|}}$$Intersecting lines: $SD = 0$ (the numerator scalar triple product vanishes).
| Case | Formula |
|---|---|
| Skew (vector) | $\dfrac{ |
| Skew (Cartesian) | $\dfrac{ |
| Parallel | $\dfrac{ |
| Intersecting | $0$ |
Before plugging into the skew formula, check the easy cases: proportional direction ratios mean parallel (use the parallel formula), and a zero numerator means intersecting ($SD = 0$). Always take the absolute value - distance is never negative.
Line-pair classification at a glance:
| Type | Intersect | Parallel | Coplanar |
|---|---|---|---|
| Intersecting | Yes | No | Yes |
| Parallel | No | Yes | Yes |
| Skew | No | No | No |
Coplanarity
Four points $A, B, C, D$ are coplanar iff the scalar triple product vanishes:
$$\boxed{[\,\vec{AB}\ \ \vec{AC}\ \ \vec{AD}\,] = 0}$$Equivalently, by determinant of differences:
$$\boxed{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \\ x_4-x_1 & y_4-y_1 & z_4-z_1 \end{vmatrix} = 0}$$Or with the $4\times 4$ determinant:
$$\boxed{\begin{vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end{vmatrix} = 0}$$Two lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ are coplanar iff:
$$\boxed{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = 0}$$Cartesian condition for two lines to be coplanar:
$$\boxed{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0}$$Normal to the plane containing two coplanar lines: $\boxed{\vec{n} = \vec{b_1} \times \vec{b_2}}$, then use the point-normal form with any point on either line.
| Item | Coplanarity condition |
|---|---|
| 4 points | $[\vec{AB}\ \vec{AC}\ \vec{AD}] = 0$ |
| 2 lines | $[(\vec{a_2}-\vec{a_1})\ \vec{b_1}\ \vec{b_2}] = 0$ |
| 3 vectors | $[\vec{u}\ \vec{v}\ \vec{w}] = 0$ |
| Plane normal (2 lines) | $\vec{n} = \vec{b_1} \times \vec{b_2}$ |