Geometry of lines and planes I

Distance of two points in 3 dimensions
$|P_1P_2|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$
$\begin{align}|P_1P_2|^2 &=|P_1B|^2+|BP_2|^2\\&=|P_1A|^2+|AB|^2+|BP_2|^2\\&=(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\end{align}$

Direction cosine
$\vec{a}=\langle a_1,a_2,a_3 \rangle$
The angles $\alpha,\beta,\gamma$ in $[0,\pi]$ that $\vec{a}$ makes with $x,y,z$ axes respectively are direction angles of $\vec{a}$.
$\Large\cos \alpha=\frac{\vec{a}\cdot \hat{i}}{|\vec{a}||\hat{i}|}=\frac{\langle a_1,a_2,a_3 \rangle \cdot \langle 1,0,0 \rangle}{|\langle a_1,a_2,a_3 \rangle||\langle 1,0,0 \rangle|}=\frac{a_1}{|\vec{a}|}$
Similarly, $\large \cos \beta=\frac{a_2}{|\vec{a}|}$ and $\large \cos \gamma=\frac{a_3}{|\vec{a}|}$.
Thus, $\vec{a}=|\vec{a}|\langle \cos\alpha,\cos\beta,\cos\gamma \rangle$
Note also that $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=\frac{1}{|\vec{a}|^2}(a_1^2+a_2^2+a_3^2)=1$

Dot product in component form
Cosine Law: $\begin{align} |\vec{a}-\vec{b}|^2&=|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|\cos \theta\\ &=|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b}\end{align}$
$\begin{align} \vec{a}\cdot \vec{b}&=|\vec{a}||\vec{b}|\cos \theta\\
&=\frac{1}{2}(|\vec{a}|^2+|\vec{b}|^2-|\vec{a}-\vec{b}|^2)\\&=\frac{1}{2}[a_1^2+a_2^2+a_3^2+b_1^2+b_2^2+b_3^2-(a_1-b_1)^2-(a_2-b_2)^2-(a_3-b_3)^2]\\&=a_1b_1+a_2b_2+a_3b_3 \end{align}$

Projections
scalar projection of $\vec{b}$ onto $\vec{a}$:
$\large \text{comp}_\vec{a} \vec{b}=|\vec{b}|\cos \theta=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}|}=\frac{\vec{a}}{|\vec{a}|}\cdot \vec{b}$
vector projection of $\vec{b}$ onto $\vec{a}$:
$\large \text{proj}_\vec{a} \vec{b}=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}|} \cdot \frac{\vec{a}}{|\vec{a}|}=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}|^2} \vec{a}$

Scalar triple product
volume of a parallelepiped $V=Ah=|\vec{b}\times\vec{c}||\vec{a}|\cos \theta=|\vec{a}\cdot (\vec{b} \times \vec{c})|$
The base parallelogram can be determined by [$\vec{b}$ and $\vec{c}$] or [$\vec{a}$ and $\vec{b}$].
Also, because of the commutative property of dot products, we have $\vec{a}\cdot (\vec{b} \times \vec{c})=(\vec{a} \times \vec{b}) \cdot \vec{c}$.

symmetric equation, vector equation, scalar equation of the plane through a point with normal vector, linear equation

skew lines

More to know [later]
Orthogonal projection
$\large \text{ortho}_\vec{a} \vec{b}=\vec{b}-\text{proj}_\vec{a} \vec{b}$

Parallelogram Law
$|\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2=2|\vec{a}|^2+2|\vec{b}|^2$
geometric meaning

Vector and chemistry
Bond angle 109.5

Vector triple product
$\vec{a}\times\vec{b}\times\vec{c}=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{a}\cdot\vec{b})\vec{c}$
$\Rightarrow \vec{a}\times(\vec{b}\times\vec{c})+\vec{b}\times(\vec{c}\times\vec{a})+\vec{c}\times(\vec{a}\times\vec{b})=\vec{0}$

$(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})=\begin{vmatrix}\vec{a}\cdot\vec{c} & \vec{b}\cdot\vec{c}\\ \vec{a}\cdot\vec{d} & \vec{b}\cdot\vec{d} \end{vmatrix}$

crystallography

Useful:
http://www.math.jhu.edu/~hhaosu/Teaching/0607FCalcIII/
http://www.math.jhu.edu/~hhaosu/Teaching/0607FCalcIII/06FMA261AHW02Sol.pdf
http://www.leadinglesson.com/problem-on-orthogonal-matrices