The points $A$ and $B$ have position vectors $$\mathbf{r}_A = \mathbf{i} + \mathbf{j} + \mathbf{k}$$ $$\mathbf{r}_B = 3\mathbf{i} - \mathbf{j} - \mathbf{k}$$ Find $\overrightarrow{AB}$.
$2\mathbf{i} - 2\mathbf{j} - 2\mathbf{k}$
The points $A$ and $B$ have position vectors $$\mathbf{r}_A = 2\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$$ $$\mathbf{r}_B = 5\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}$$ Find $|\overrightarrow{AB}|$.
$\sqrt{74}$
The points $A$ and $B$ have position vectors $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}$ and $\mathbf{b} = 5\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ respectively. - Find the vector $\overrightarrow{AB}$.
- Find the unit vector in the direction of $\overrightarrow{AB}$.
- Find the position vector of the midpoint $M$ of $AB$.
(a) $3\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}$ $\quad$ (b) $\dfrac{1}{\sqrt{34}}(3\mathbf{i} - 4\mathbf{j} + 3\mathbf{k})$ $\quad$ (c) $\dfrac{7}{2}\mathbf{i} + \mathbf{j} + \dfrac{1}{2}\mathbf{k}$ - $\overrightarrow{AB} = 3\mathbf{i} - 4\mathbf{j} + 3\mathbf{k}$
- $\hat{u} = \dfrac{1}{\sqrt{34}}(3\mathbf{i} - 4\mathbf{j} + 3\mathbf{k})$
- $\mathbf{m} = \dfrac{7}{2}\mathbf{i} + \mathbf{j} + \dfrac{1}{2}\mathbf{k}$
The points $A$, $B$ and $C$ have position vectors $$\mathbf{r}_A = 3\mathbf{i} + 4\mathbf{j} - \mathbf{k}$$ $$\mathbf{r}_B = \mathbf{i} + \mathbf{j} + \mathbf{k}$$ $$\mathbf{r}_C = 5\mathbf{i} + 7\mathbf{j} - 3\mathbf{k}$$ Determine if the poitns are collinear.
$\overrightarrow{AB} = -\overrightarrow{AC}$ and they share the point $A$, so yes.
The points $A$, $B$ and $C$ are collinear and have position vectors $$\mathbf{r}_A = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$$ $$\mathbf{r}_B = 3\mathbf{i} - \mathbf{j} + \mathbf{k}$$ $$\mathbf{r}_C = 5\mathbf{i} + p\mathbf{j} + q\mathbf{k}$$ Find $p$ and $q$.
$p = -4$ and $q = 5$
The points $A$ and $B$ have position vectors $$\mathbf{r}_A = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$$ $$\mathbf{r}_B = \mathbf{i} + \mathbf{j} - \mathbf{k}$$ Show that the triangle $OAB$ is right angled.
Pythagoras
The points $A$ $B$ and $C$ have position vectors $$\mathbf{r}_A = -6\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}$$ $$\mathbf{r}_B = -3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$$ $$\mathbf{r}_C = -8\mathbf{i} + 1\mathbf{j} - 6\mathbf{k}$$ Show that the triangle $ABC$ is right angled.
Pythagoras
The points $A$, $B$ and $C$ have position vectors $$\mathbf{r}_A = \mathbf{i} + \mathbf{j} - \mathbf{k}$$ $$\mathbf{r}_B = 2\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}$$ $$\mathbf{r}_C = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$$ For the triangle $ABC$, find, to 3 significant figures, - the perimeter
- the area.
The points $A$, $B$ and $C$ have position vectors $$\mathbf{r}_A = 4\mathbf{i} - \mathbf{j} - 2\mathbf{k}$$ $$\mathbf{r}_B = \mathbf{i} - 4\mathbf{j} + 2\mathbf{k}$$ $$\mathbf{r}_C = -3\mathbf{i} + 2\mathbf{j} + \mathbf{k}$$ For the triangle $ABC$, find, to 3 significant figures, - the perimeter
- the area.
The points $A$ $B$ and $C$ have position vectors $$\mathbf{r}_A = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$$ $$\mathbf{r}_B = 5\mathbf{i} + 2\mathbf{j} + 7\mathbf{k}$$ $$\mathbf{r}_C = \mathbf{i} + 4\mathbf{j} + 6\mathbf{k}$$ Given $ABCD$ is a parallelogram, find the position vector of $D$.
$-2\mathbf{i} + \mathbf{j} + 2\mathbf{k}$
The point $A$ has the two dimensional position vector $\mathbf{r} = 6\mathbf{i}$. Given $OAB$ is an equilateral triangle, find the possible position vectors for $B$.
$3\mathbf{i} + 3\sqrt{3}\mathbf{j}$ or $3\mathbf{i} - 3\sqrt{3}\mathbf{j}$
The forces $\mathbf{F}_1$ and $\mathbf{F}_2$ act on a particle. $$\mathbf{F}_1 = 2\mathbf{i} - \mathbf{j}$$ $$\mathbf{F}_2 = -3\mathbf{i} + 5\mathbf{j}$$ Find the magnitude and direction of the resultant force.
$\sqrt{17}$, $104^{\circ}$
The forces $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ act on a particle. $$\mathbf{F}_1 = 4\mathbf{i} - 2\mathbf{j}$$ $$\mathbf{F}_2 = 2\mathbf{i} + p\mathbf{j}$$ $$\mathbf{F}_3 = q\mathbf{i} - 3\mathbf{j}$$ The particle is in equilibrium. Find $p$ and $q$.
$p = 5$ and $q = -6$
The forces $\mathbf{F}_1$ and $\mathbf{F}_2$ act on a particle Given $\mathbf{F}_1 = 5\mathbf{i} - 3\mathbf{j}$ and that the resultant force has magnitude $\sqrt{8}$ and direction $-45^{\circ}$, find $\mathbf{F}_2$.
$-3\mathbf{i} + \mathbf{j}$
The forces $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ act on a particle. $$\mathbf{F}_1 = \mathbf{j}$$ $$\mathbf{F}_2 = 2\mathbf{i} + 3\mathbf{j}$$ $$\mathbf{F}_3 = p\mathbf{i} + q\mathbf{j}$$ The resultant force has magnitude $\sqrt{18}$ and direction $135^{\circ}$. Find $p$ and $q$.
$p = -5$ and $q = -1$
A particle of mass 500 g is being acted on by a force $\mathbf{F} = \mathbf{i} + \mathbf{j}$. Find the acceleration of the particle.
$2\mathbf{i} + 2\mathbf{j}$
A particle of mass 2 kg is being acted on by a force $\mathbf{F} = -4\mathbf{i} + 2\mathbf{j}$. Find the magnitude and direction of the acceleration of the particle.
$\sqrt{5}$, $153^{\circ}$