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Notation
In print, vectors are typeset in bold, like $\mathbf{a}$. In writing, we usually put a tilde underneath, like $\underset{\sim}{a}$.
$\overrightarrow{AB}$ is the vector represented by the directed line segment $AB$.
$| \textbf{ a }|$: the magnitude of $\mathbf{a}$.
$\hat{\mathbf{a}}$: a unit vector in the direction of $\mathbf{a}$.
$\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \, \mathbf{j} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \, \mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.
$\overrightarrow{AB}$ is the vector represented by the directed line segment $AB$.
$| \textbf{ a }|$: the magnitude of $\mathbf{a}$.
$\hat{\mathbf{a}}$: a unit vector in the direction of $\mathbf{a}$.
$\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \, \mathbf{j} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, \, \mathbf{k} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$.
Ratio theorem formula (provided)
The point dividing $AB$ in the ratio $\lambda:\mu$ has position vector $ \displaystyle \frac{\mu \mathbf{a} + \lambda \mathbf{b}}{\lambda + \mu}$
$ \overrightarrow{OA} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix},
\overrightarrow{OB} = \begin{pmatrix} 0 \\ 4 \\ -3 \end{pmatrix} $.
Find $\overrightarrow{OC}$, given that $OACB$ is a parallelogram.
Find $\overrightarrow{OC}$, given that $OACB$ is a parallelogram.
$\overrightarrow{OC}= \begin{pmatrix} 2 \\ 1 \\ -2 \end{pmatrix}$
Given that $ \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$ is parallel to $\begin{pmatrix} -3 \\ y \\ z \end{pmatrix}$, find $y$ and $z$.
$y=-7, z=17$
Explain why $\begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$ is parallel to $\begin{pmatrix} 4 \\ 8 \\ -12 \end{pmatrix}$.
Given that $ \mathbf{a} = \mathbf{i} - \mathbf{j} + 2\mathbf{k}$, find $| \mathbf{a} |$ and $\hat{\mathbf{a}}$.
$\sqrt{6}, \frac{1}{\sqrt{6}} (\mathbf{i} - \mathbf{j} + 2 \mathbf{k})$
Find a vector that is of magnitude $5$ and is parallel to $\mathbf{a}$.
$\frac{5}{\sqrt{6}}(\mathbf{i} - \mathbf{j} + 2 \mathbf{k}) $
Given points $A (5,-1,3)$ and $B$ with position vector $\overrightarrow{OB} = \mathbf{i} + 2 \mathbf{j} - 4 \mathbf{k}$, find the vector $\overrightarrow{AB}$.
$(-4\mathbf{i} +3 \mathbf{j} - 7\mathbf{k})$
Find the distance between $A$ and $B$.
$\sqrt{74} \textrm{ units}$
It is further given that $C$ has coordinates $(13, y, 17)$. Given that $A,B$ and $C$ are collinear, find $y$. $y=-7$
If $D$ has coordinates $(-3, 5, -11)$. Show that $A,B$ and $D$ lie on the same line.
The points $A$ and $B$ have coordinates $(1, 2, -3)$ and $(-2,0,5)$ respectively.
Given that the point $P$ divides the line $AB$ in the ratio $1:2$, find the position vector of $P$. $\frac{4}{3} \mathbf{j} - \frac{1}{3} \mathbf{k}$
Given that the point $P$ divides the line $AB$ in the ratio $1:2$, find the position vector of $P$. $\frac{4}{3} \mathbf{j} - \frac{1}{3} \mathbf{k}$
It is given that the point $Q$ lies on $AB$ extended such that $AB:AQ=2:5$.
Find the coordinates of $Q$. $\left (-\frac{13}{2}, -3, 17 \right )$
Find the coordinates of $Q$. $\left (-\frac{13}{2}, -3, 17 \right )$
Let $\mathbf{a}$ and $\mathbf{b}$ be non-parallel vectors that are non-zero.
It is given that $\overrightarrow{OX} = \lambda \mathbf{a} + \frac{1-\lambda}{2} \mathbf{b}$ and $\overrightarrow{OY} = \frac{1}{2} \mathbf{a} + \frac{1}{2} \mathbf{b}$,
where $\lambda$ is a real constant, $\lambda \neq 0$.
Given that $O,X$ and $Y$ are collinear, find the value of the $\lambda$. $\lambda = \frac{1}{3}$
It is given that $\overrightarrow{OX} = \lambda \mathbf{a} + \frac{1-\lambda}{2} \mathbf{b}$ and $\overrightarrow{OY} = \frac{1}{2} \mathbf{a} + \frac{1}{2} \mathbf{b}$,
where $\lambda$ is a real constant, $\lambda \neq 0$.
Given that $O,X$ and $Y$ are collinear, find the value of the $\lambda$. $\lambda = \frac{1}{3}$