Three-Dimensional (3D) Geometry

Spatial Coordinate System

Points in ${\mathbb{R}}^3$ are defined by triples $(x,y,z)$. Direction cosines $(l,m,n)$ are the cosines of angles made with the positive $x,y,z$ axes, satisfying $l^2+m^2+n^2=1$.

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1. Straight Line in 3D Space

📊Visual Mathematical Intuition

3D Spatial Geometry: Skew Lines Shortest Distance & Normal Plane

Skew Lines Distance:$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$

Plane Equation:$A(x - x_1) + B(y - y_1) + C(z - z_1) = 0 \iff \vec{r} \cdot \vec{n} = d$

Direction Cosines:$l^2 + m^2 + n^2 = 1, \; \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

3D skew lines $L_1, L_2$ separated by their unique common perpendicular shortest distance segment $d$, projected along the common normal vector $\vec{n} = \vec{b}_1 \times \vec{b}_2$.

• Vector Equation: $\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}$
• Cartesian (Symmetric) Equation: $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Shortest Distance Between Two Skew Lines

Given $L_1:\overrightarrow{r}={\overrightarrow{a}}_1+\lambda {\overrightarrow{b}}_1$ and $L_2:\overrightarrow{r}={\overrightarrow{a}}_2+\mu {\overrightarrow{b}}_2$:

$$d=\frac{|({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)\cdot ({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)|}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}$$

• Condition for Coplanarity / Intersection:$$({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)\cdot ({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2)=0⟺\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$$

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2. Equation of a Plane in 3D Space

• Vector Equation: $\overrightarrow{r}\cdot \overrightarrow{n}=d$
• Cartesian Form: $Ax+By+Cz+D=0$ (Normal vector $\overrightarrow{n}=A\hat{i}+B\hat{j}+C\hat{k}$)
• Plane Passing Through $(x_1,y_1,z_1)$ with Normal $(A,B,C)$:$$A(x-x_1)+B(y-y_1)+C(z-z_1)=0$$
• Plane Passing Through 3 Non-Collinear Points:$$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0$$

Perpendicular Distance from $(x_1,y_1,z_1)$ to Plane $Ax+By+Cz+D=0$

$$p=\frac{|A{x}_1+B{y}_1+C{z}_1+D|}{\sqrt{A^2+B^2+C^2}}$$

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3. Intersection of Line and Plane

For line $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ and plane $Ax+By+Cz+D=0$:

• Line is parallel to plane $⟺Aa+Bb+Cc=0$.
• Line lies entirely in plane $⟺Aa+Bb+Cc=0$ AND $A{x}_1+B{y}_1+C{z}_1+D=0$.
• Angle $\theta$ between line and plane:$$\sin \theta =\frac{|Aa+Bb+Cc|}{\sqrt{a^2+b^2+c^2}\sqrt{A^2+B^2+C^2}}$$