3D Geometry Formulas
Chapter 11: CBSE 2025-26 Compilation
1. Direction Cosines & Ratios
| Concept | Formula |
|---|---|
| Direction Cosines (l, m, n) | $$ l=\cos\alpha, \quad m=\cos\beta, \quad n=\cos\gamma $$ |
| Fundamental Identity | $$ l^2 + m^2 + n^2 = 1 $$ |
| From Direction Ratios (a, b, c) | $$ l = \frac{a}{\sqrt{a^2+b^2+c^2}} $$ (Similarly for $$ m $$ and $$ n $$) |
2. Equations of Straight Lines
| Form | Equation |
|---|---|
| Vector Form (Point $$ \vec{a} $$, Parallel to $$ \vec{b} $$) |
$$ \vec{r} = \vec{a} + \lambda\vec{b} $$ |
| Cartesian Form (Point $$ x_1, y_1, z_1 $$, DRs $$ a, b, c $$) |
$$ \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} $$ |
| Two Point Form (Through $$ P(x_1…) $$ & $$ Q(x_2…) $$) |
$$ \frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1} $$ |
3. Angles & Conditions
Parallel Condition:
$$ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $$
$$ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} $$
Perpendicular Condition:
$$ a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 $$
$$ a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 $$
| Angle ($$ \theta $$) between two lines | $$ \cos\theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{\sum a_1^2}\sqrt{\sum a_2^2}} \right| $$ Vector: $$ \cos\theta = \left| \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|} \right| $$ |
4. Shortest Distance (Skew & Parallel)
A. Distance between Skew Lines ($$ \vec{r} = \vec{a}_1 + \lambda\vec{b}_1 $$ & $$ \vec{r} = \vec{a}_2 + \mu\vec{b}_2 $$)
Vector Form:
$$ d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right| $$
$$ d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 – \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right| $$
Cartesian Form:
$$ d = \frac{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}}{\sqrt{(b_1 c_2 – b_2 c_1)^2 + (c_1 a_2 – c_2 a_1)^2 + (a_1 b_2 – a_2 b_1)^2}} $$
B. Distance between Parallel Lines ($$ \vec{r} = \vec{a}_1 + \lambda\vec{b} $$ & $$ \vec{r} = \vec{a}_2 + \mu\vec{b} $$)
$$ d = \left| \frac{\vec{b} \times (\vec{a}_2 – \vec{a}_1)}{|\vec{b}|} \right| $$