Properties & Formulas

Column vector and i, j, k notations

The position vector of point $P(2, 2, 3)$ in column vector form: $$ \overrightarrow{OP} = \left( \begin{matrix} 2 \\ 2 \\ 3 \end{matrix} \right) $$

$\textbf{i}$ represents a unit vector in the $x$-direction, i.e. $\textbf{i} = \left( \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right)$

$\textbf{j}$ represents a unit vector in the $y$-direction, i.e. $\textbf{j} = \left( \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right)$

$\textbf{k}$ represents a unit vector in the $z$-direction, i.e. $\textbf{k} = \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right)$

Thus, the position vector of point $P(2, 2, 3)$ in $\textbf{i}$, $\textbf{j}$, $\textbf{k}$ notations: $$ \overrightarrow{OP} = 2\textbf{i} + 2 \textbf{j} + 3 \textbf{k} $$

Vector subtraction

Finding the displacement vector $\overrightarrow{AB}$ from vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$: $$ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} $$

Magnitude of vector

For a 2D vector like $\overrightarrow{AB} = \left( \begin{matrix} x \\ y \end{matrix} \right)$, the magnitude of $\overrightarrow{AB}$ is given by $$ \boxed{ \left| \overrightarrow{AB} \right| = \sqrt{x^2 + y^2} } $$

For a 3D vector like $\textbf{c} = \left( \begin{matrix} x \\ y \\ z \end{matrix} \right)$, the magnitude of $\textbf{c}$ is given by $$ \boxed{ | \textbf{c} | = \sqrt{x^2 + y^2 + z^2 } } $$

Unit vector

The unit vector of vector $\textbf{a}$ is denoted by $ \hat{ \textbf{a} } $ and can be found by: $$ \boxed{ \hat{ \textbf{a} } = {1 \over | \textbf{a} | } \textbf{a} } $$

The magnitude of a unit vector is equals to 1.

Parallel vectors

Vectors $\textbf{a}$ and $\textbf{b}$ are parallel if $$ \boxed{ \textbf{a} = k \textbf{b}, \text{ where } k \in \mathbb{R} } $$

For example, the vectors $ \textbf{c} = \left( \begin{matrix} 1 \\ 0 \\ 2 \end{matrix} \right) $ and $ \textbf{d} = \left( \begin{matrix} -2 \\ 0 \\ -4 \end{matrix} \right) $ are parallel since \begin{align} \textbf{d} & = \left( \begin{matrix} -2 \\ 0 \\ -4 \end{matrix} \right) \\ \textbf{d} & = -2 \left( \begin{matrix} 1 \\ 0 \\ 2 \end{matrix} \right) \\ \textbf{d} & = -2 \textbf{c} \end{align}

Collinearity

Points $A$, $B$ and $C$ collinear if they lie on a straight line

To prove that the three points are collinear, show that $$ \boxed{ \overrightarrow{AB} = k \overrightarrow{BC}, \text{ where } k \in \mathbb{R} } $$

Ratio theorem

If the point $M$ divides the line segment $AB$ in the ratio $\lambda : \mu$, the position vector of $M$ is given by $$ \boxed{ \overrightarrow{OM} = { \mu \overrightarrow{OA} + \lambda \overrightarrow{OB} \over \lambda + \mu } } $$