Vectors in 2D (AQA A Level Maths: Pure): Exam Questions

[In this question, $\mathbf{i}$ is a unit vector due east and $\mathbf{j}$ is a unit vector due north.]

A ship leaves a port at the fixed origin $O$ and travels $300$ km on a bearing of $120°$. It then travels $500$ km due south before dropping anchor at the point $A$.

The position vector of $A$ relative to $O$ is $(x\mathbf{i}+y\mathbf{j})$ km.

Find the exact value of $x$ and the exact value of $y$.

Two forces, ${\mathbf{F}}_1$ and ${\mathbf{F}}_2$, act on a particle, where

${\mathbf{F}}_1=(7\mathbf{i}-2\mathbf{j})\text{ N}$

${\mathbf{F}}_2=(-12\mathbf{i}-10\mathbf{j})\text{ N}$

The resultant of these two forces is $\mathbf{R}$.

Find the magnitude of $\mathbf{R}$.

A third force, ${\mathbf{F}}_3=k\mathbf{j}$ N, where $k$ is a constant, is now applied to the particle.

Given that the new resultant of the three forces acts at an angle of $45°$ to the positive $\mathbf{j}$ direction, measured anticlockwise, find the value of $k$.

Relative to a fixed origin $O$, the points $A$, $B$ and $C$ have position vectors

$\overrightarrow{OA}=-4\mathbf{i}-7\mathbf{j}$

$\overrightarrow{OB}=3\mathbf{j}$

$\overrightarrow{OC}=6\mathbf{i}+18\mathbf{j}$

Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

Show that $\overrightarrow{AB}$ is parallel to $\overrightarrow{AC}$, and hence state what this tells you about the points $A$, $B$ and $C$.

Figure 2 shows a sketch of triangle $ABC$.

Given that

$\overrightarrow{AB}=\mathbf{a}$

$\overrightarrow{AC}=\mathbf{b}$

and that the point $P$ lies on $BC$ such that $BP:PC=3:2$,

(i) find $\overrightarrow{BC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$,

(ii) hence find $\overrightarrow{BP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Given that $\mathbf{a}=7\mathbf{i}+8\mathbf{j}$ and $\mathbf{b}=12\mathbf{i}+3\mathbf{j}$, find $\overrightarrow{BP}$ in terms of $\mathbf{i}$ and $\mathbf{j}$.

Figure 1 shows a sketch of triangle $ABC$.

Given that

$\overrightarrow{AB}=7\mathbf{i}+\mathbf{j}$

$\overrightarrow{AC}=4\mathbf{i}-3\mathbf{j}$

(i) Write down $\overrightarrow{CA}$ in terms of $\mathbf{i}$ and $\mathbf{j}$.

(ii) Find $\overrightarrow{BC}$ in terms of $\mathbf{i}$ and $\mathbf{j}$.

Find the exact value of $\left|\overrightarrow{BC}\right|$.

The vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are given by

$\mathbf{a}=\left(\begin{array}{c}7\\ 2\end{array}\right),\mathbf{b}=\left(\begin{array}{c}m\\ -3\end{array}\right),\mathbf{c}=\left(\begin{array}{c}5\\ n\end{array}\right)$

where $m$ and $n$ are constants.

Given that $\mathbf{a}+2\mathbf{b}=\mathbf{c}$, find the value of $m$ and the value of $n$.

The vector $\mathbf{d}$ is given by

$\mathbf{d}=\left(\begin{array}{c}-5\\ k\end{array}\right)$

where $k$ is a constant.

Given that $\left|\mathbf{d}\right|=15$, find the two possible values of $k$, giving your answers as simplified surds.

Relative to a fixed origin $O$, the point $A$ has position vector $2k\mathbf{i}+7k\mathbf{j}$, where $k$ is a constant.

The point $A$ lies on the straight line $l$ with equation

$y=3x+5$

Find the value of $k$, and hence determine the coordinates of $A$.

The vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are given by

$\mathbf{a}=\left(\begin{array}{c}-5\\ 17\end{array}\right),\mathbf{b}=\left(\begin{array}{c}k\\ 5\end{array}\right),\mathbf{c}=\left(\begin{array}{c}9\\ -29\end{array}\right)$

where $k$ is a constant.

Given that $\mathbf{a}-\mathbf{b}$ is parallel to $\mathbf{b}+\mathbf{c}$, find the value of $k$.

The vector $\overrightarrow{AB}$ is given by

$\overrightarrow{AB}=11\mathbf{i}-2\mathbf{j}$

Find

(i) the exact value of $\left|\overrightarrow{AB}\right|$,

(ii) the angle between $\overrightarrow{AB}$ and the positive $x$-axis, giving your answer in degrees to two decimal places.

Find a unit vector in the direction of $\overrightarrow{AB}$, giving your answer in its simplest exact form.

Figure 1 shows a sketch of triangle $ABC$.

Given that

$\overrightarrow{AB}=5\mathbf{i}+8\mathbf{j}$

$\overrightarrow{BC}=\mathbf{i}-5\mathbf{j}$

Explain geometrically why $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\mathbf{0}$.

Find $\overrightarrow{CA}$ and hence find the exact value of $\left|\overrightarrow{CA}\right|$.

The vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are given by

$\mathbf{a}=\left(\begin{array}{c}-1\\ n\end{array}\right),\mathbf{b}=\left(\begin{array}{c}5\\ -4\end{array}\right),\mathbf{c}=\left(\begin{array}{c}m\\ 6\end{array}\right)$

where $m$ and $n$ are constants.

Given that the resultant of $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ is the zero vector, find the value of $m$ and the value of $n$.

The vector $\mathbf{d}$ is given by

$\mathbf{d}=\left(\begin{array}{c}-3k\\ k\end{array}\right)$

where $k$ is a constant.

Given that $\left|\mathbf{d}\right|=2\sqrt{15}$, find the two possible values of $k$, giving your answers as simplified surds.

Relative to a fixed origin $O$, the point $A$ has position vector $3k\mathbf{i}-17k\mathbf{j}$, where $k$ is a positive constant.

The point $A$ lies on the curve $C$ with equation

$y=x^2-2$

Find the value of $k$, and hence determine the coordinates of $A$.

The vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are given by

$\mathbf{a}=\left(\begin{array}{c}3\\ 5\end{array}\right),\mathbf{b}=\left(\begin{array}{c}-3k\\ k\end{array}\right),\mathbf{c}=\left(\begin{array}{c}0\\ -4\end{array}\right)$

where $k$ is a constant.

Given that $\mathbf{a}-\mathbf{b}$ is parallel to $\mathbf{a}+\mathbf{c}$, find the value of $k$.

The vector $\overrightarrow{AB}$ has magnitude $6\sqrt{3}$ and makes an angle of $150°$ with the positive $x$-axis, measured anticlockwise.

Find $\overrightarrow{AB}$ in the form $x\mathbf{i}+y\mathbf{j}$, where $x$ and $y$ are exact constants to be found.

Find a unit vector in the direction of $\overrightarrow{AB}$, giving your answer in its simplest exact form.

[In this question, $\mathbf{i}$ is a unit vector due east and $\mathbf{j}$ is a unit vector due north. Position vectors are given relative to a fixed origin $O$.]

In the enchanted kingdom of Vectoria, a magical flying unicorn takes off from the wizard's palace at the point $O$ and travels $30$ km on a bearing of $300°$.

Chased by an evil dragon, it then travels an unknown distance of $k$ km due north before reaching the enchanted grove at the point $P$, where $k$ is a positive constant.

The position vector of $P$ relative to $O$ is $(x\mathbf{i}+y\mathbf{j})$ km.

Given that the straight-line distance between the grove and the palace is $30\sqrt{3}$ km, find the exact value of $x$ and the exact value of $y$.

[In this question the unit vectors $\mathbf{i}$ and $\mathbf{j}$ are due east and due north respectively.]

A rescue helicopter leaves a fixed origin $O$ and flies in the direction of a lost hiker. The helicopter flies at a constant speed of 120 km h⁻¹ in a direction parallel to the vector $3\mathbf{i}+4\mathbf{j}$.

After flying for 15 minutes, the helicopter reaches the point $P$.

From P, the helicopter immediately changes direction and flies on a bearing of 270° at the same constant speed.

After a further 20 minutes, the helicopter reaches the hiker at the point $Q$.

Given that the position vector of $Q$ relative to $O$ is $\left(x\mathbf{i}+y\mathbf{j}\right)\mathrm{km}$,

find the exact values of $x$ and $y$.

Find the distance and the bearing of the hiker from $O$. Give the distance in km and the bearing in degrees, each to one decimal place.

Two forces ${\mathbf{F}}_1$ and ${\mathbf{F}}_2$ act on a particle, where

${\mathbf{F}}_1=(5\mathbf{i}-3\mathbf{j})\text{ N}$

${\mathbf{F}}_2=(x\mathbf{i}+y\mathbf{j})\text{ N}$

where $x$ and $y$ are scalar constants.

The resultant force $\mathbf{R}$ acting on the particle is given by $\mathbf{R}={\mathbf{F}}_1+{\mathbf{F}}_2$.

Given that $\mathbf{R}$ acts in a direction parallel to the vector $(-\mathbf{i}-3\mathbf{j})$, find the angle between $\mathbf{R}$ and the vector $\mathbf{j}$, giving your answer in degrees to two decimal places.

Show that $3x-y=-18$.

Given that $y=-3$, find the magnitude of $\mathbf{R}$.

Relative to a fixed origin $O$, the points $A$, $B$ and $C$ have position vectors

$\overrightarrow{OA}=-9\mathbf{i}+4\mathbf{j}$

$\overrightarrow{OB}=-6\mathbf{i}$

$\overrightarrow{OC}=3\mathbf{i}-12\mathbf{j}$

Show, using a vector method, that the points $A$, $B$ and $C$ lie on a straight line.

Figure 2 shows a sketch of triangle $ABC$.

The point $F$ lies on $AB$ such that $AF:FB=m:n$, where $m$ and $n$ are positive constants.

The point $G$ lies on the line segment $BC$.

The line segment $FG$ is parallel to $AC$.

Explain why $\overrightarrow{BG}=\lambda \overrightarrow{BC}$ for some scalar constant $\lambda$, where $0<\lambda <1$.

Given that

$\overrightarrow{AB}=\mathbf{a},\overrightarrow{AC}=\mathbf{b}$

show that

$\overrightarrow{FG}=\left(\frac{n}{m+n}-\lambda \right)\mathbf{a}+\lambda \mathbf{b}$

Hence prove that $BG:GC=n:m$.

The points $A$, $B$ and $C$ are the vertices of a triangle.

Given that

$\overrightarrow{AC}=5\mathbf{i}-2\mathbf{j}$

$\overrightarrow{BC}=-3\mathbf{i}+k\mathbf{j}$

where $k$ is a constant,

Find $\overrightarrow{AB}$ in terms of $\mathbf{i}$, $\mathbf{j}$ and $k$.

Given that $\left|\overrightarrow{AB}\right|=\sqrt{89}$, find the two possible values of $k$.

The vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are given by

$\mathbf{a}=\left(\begin{array}{c}8\\ m\end{array}\right),\mathbf{b}=\left(\begin{array}{c}n\\ -2\end{array}\right),\mathbf{c}=\left(\begin{array}{c}m\\ n\end{array}\right)$

where $m$ and $n$ are constants.

Given that $\mathbf{a}+\mathbf{b}=\mathbf{c}-2\mathbf{b}$, find the value of $m$ and the value of $n$.

The vector $\mathbf{d}$ is given by

$\mathbf{d}=\left(\begin{array}{c}2k+1\\ 2k-1\end{array}\right)$

where $k$ is a constant.

Given that $\left|\mathbf{d}\right|=3k\sqrt{2}$, find the two possible values of $k$, giving your answers in simplest exact form.

Relative to a fixed origin $O$, the point $A$ has position vector $3k\mathbf{i}+5k\mathbf{j}$, where $k$ is a constant.

The point $A$ lies on the circle $C$ with equation

$(x-11{)}^2+(y-7{)}^2=34$

Find the value of $k$, and hence determine the coordinates of $A$.

The straight line $l$ passes through $O$ and $A$.

Explain algebraically why $l$ must be a tangent to $C$.

Relative to a fixed origin $O$, the points $A$, $B$ and $C$ have position vectors

$\overrightarrow{OA}=-6\mathbf{i}-2\mathbf{j}$

$\overrightarrow{OB}=\mathbf{i}+m\mathbf{j}$

$\overrightarrow{OC}=3\mathbf{i}-8\mathbf{j}$

where $m$ is a constant.

Given that $A$, $B$ and $C$ lie on the same straight line, use a vector method to find the exact value of $m$.

The vector $\overrightarrow{AB}$ has magnitude $2\sqrt{6}$ and makes an angle of $165°$ with the positive $y$-axis, measured anticlockwise.

Find $\overrightarrow{AB}$ in the form $a\mathbf{i}+b\mathbf{j}$, where $a$ and $b$ are exact constants to be found.

Find a unit vector in the direction of $\overrightarrow{AB}$, giving your answer in its simplest exact form.

[In this question, $\mathbf{i}$ is a unit vector due east and $\mathbf{j}$ is a unit vector due north. Position vectors are given relative to a fixed origin $O$.]

A ship searches for a radio buoy. The ship sets out from $O$ and moves with a constant speed of $40$ km/h in a direction parallel to the vector $\mathbf{i}+3\mathbf{j}$.

After $90$ minutes, the ship reaches the point $P$. At $P$, the ship receives a transmission indicating that the buoy is on a bearing of $210°$ from the ship.

The ship immediately changes course and travels on a bearing of $210°$ at a constant speed of $40$ km/h for a further $45$ minutes, reaching the buoy at the point $Q$.

Given that the position vector of $Q$ relative to $O$ is $(x\mathbf{i}+y\mathbf{j})$ km, find the exact value of $x$ and the exact value of $y$.

Find the distance of the buoy from the ship, and its bearing from the ship, at the time the ship initially left $O$.

Give your distance in km to one decimal place, and your bearing to one decimal place.

Three forces, ${\mathbf{F}}_1$, ${\mathbf{F}}_2$ and ${\mathbf{F}}_3$, act on a particle, where

${\mathbf{F}}_1=(7\mathbf{i}-\mathbf{j})\text{ N}$

${\mathbf{F}}_2=(x\mathbf{i}+y\mathbf{j})\text{ N}$

${\mathbf{F}}_3=(k\mathbf{i}+k\sqrt{3} \mathbf{j})\text{ N}$

where $x$ and $y$ are scalar constants, and $k$ is a positive constant.

The resultant force acting on the particle is $\mathbf{R}$.

Given that $\mathbf{R}=\mathbf{0}$ when $\left|{\mathbf{F}}_3\right|=10$ N, find the exact value of $x$ and the exact value of $y$.

Find the magnitude of ${\mathbf{F}}_2$ and the angle it makes with the positive $\mathbf{i}$ direction. Give both answers correct to one decimal place.

Figure 3 shows a sketch of triangle $ABC$.

The point $D$ is the midpoint of $AB$. The point $E$ is the midpoint of $AC$. The line segments $BE$ and $CD$ intersect at the point $F$.

Given that $\overrightarrow{AB}=2\mathbf{a}$ and $\overrightarrow{AC}=2\mathbf{b}$, find $\overrightarrow{BC}$, $\overrightarrow{BE}$ and $\overrightarrow{CD}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

By setting up and solving suitable vector equations, prove that each of the line segments $BE$ and $CD$ divides the other in the ratio $1:2$.