Vectors in 3 Dimensions (Cambridge (CIE) A Level Maths: Pure 3): Exam Questions

The coordinates of points A and B are $\left(-2,5,-7\right)$ and $\left(-7,k,3\right)$ respectively.  Given that the distance from A to B is $5\sqrt{14}$ units, find the possible values of k.

In the triangle ABC,  $\overrightarrow{AB}=-2\mathbf{i}+3\mathbf{j}-\mathbf{k}$ and  $\overrightarrow{AC}=-5\mathbf{i}-4\mathbf{j}-7\mathbf{k}$.

Show that $\angle BAC=81.9°$ to 1 d.p.

Point R has coordinates $\left(-1,5,14\right)$ and point S has coordinates$\left(7,-2,12\right)$.

Find:

(i) the vector $\overrightarrow{RS}$,

(ii) the unit vector in the direction of $\overrightarrow{RS}$.

Find the angle $\overrightarrow{RS}$ makes with the positive y-axis.

The vector  $\overrightarrow{TU}=-24\mathbf{i}+21\mathbf{j}+6\mathbf{k}.$

Explain, giving a reason for your answer, whether the vectors $\overrightarrow{\mathit{R}\mathit{S}}$ and $\overrightarrow{TU}$ are parallel.

The triangle PQR has vertices $P\left(12,3,-3\right)$, $Q\left(7,-8,k\right)$and $R\left(3,3,-12\right)$.  Given that $PQR$ is an equilateral triangle, find the value of k.

Vectors a and b are defined by 

$\begin{array}{rcl}\mathbf{a}& =& -12\mathbf{i}-7\mathbf{j}+15\mathbf{k}\\ \mathbf{b}& =& 4p\mathbf{i}+\left(pqr+2qr-p\right)\mathbf{j}-pq\mathbf{k}\end{array}$

Given that $\mathbf{a}=\mathbf{b}$, find the values of p, q and r.

A particle of mass 0.4 kg is acted upon by a force of $\left(-2\mathbf{i}+6\mathbf{j}+10\mathbf{k}\right)\mathrm{N}$. 

Find:

(i) the acceleration of the particle while the force acts,

(ii) the magnitude of the acceleration to 3 s.f.

A second force with a magnitude of $10\sqrt{3}\mathrm{N}$ now acts on the particle. The resultant of the two forces is parallel to the vector j and has a magnitude of 8 N.

Find this second force, giving your answer in the form $\left(x\mathbf{i}+y\mathbf{j}+z\mathbf{k}\right)\mathrm{N}$.

Three forces act upon a particle with a mass of 5 kg:

$\begin{array}{rcl}{\mathbf{F}}_1& =& \left(3\mathbf{i}-7\mathbf{j}+\mathrm{p}\mathbf{k}\right)\mathrm{N}\\ {\mathbf{F}}_2& =& \left(\mathrm{q}\mathbf{i}+3\mathbf{j}-\mathbf{k}\right)\mathrm{N}\\ {\mathbf{F}}_3& =& \left(-2\mathbf{i}+\mathrm{r}\mathbf{j}-5\mathbf{k}\right)\mathrm{N}\end{array}$

Under the action of those three forces, the particle is in equilibrium.

Find the values of p, q and r.

The third force is doubled, so that the three forces now acting on the particle are F₁, F₂ and 2F₃. 

Work out:

(i) the resultant force R now acting on the particle.

(ii) the acceleration of the particle under the action of R.

(iii) the magnitude of the acceleration found in (ii), giving your answer as an exact value.

The diagram below shows a cube whose vertices are O, A, B, C, D, E, F and G.

a, b and c are the vectors $\overrightarrow{OA},\overrightarrow{OB}$ and $\overrightarrow{OC}$ respectively.

Find vectors $\overrightarrow{OE}$ and $\overrightarrow{AG}.$

Let P be a point on OE, and let Q be a point on AG.

Explain why the vectors $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ can be expressed in the forms 

$\begin{array}{rcl}\overrightarrow{OP}& =& \lambda \overrightarrow{OE}\\ \overrightarrow{OQ}& =& \mathbf{a}+\mu \overrightarrow{AG}\end{array}$

where $\lambda$ and $\mu$  are constants with $0\le \lambda \le 1$ and $0\le \mu \le 1$ 

By solving the equation $\overrightarrow{OP}=\overrightarrow{OQ}$, using your results from (a) and (b), show that the diagonals OE and AG intersect each other, and determine the ratios into which they are cut by the point of intersection.

The coordinates of points A and B are $\left(-1,3,14\right)$and $\left(2k,3k,13\right)$ respectively.
Given that the distance from A to B is $\sqrt{163}$units, and that k is an integer, find the value of k.

In the triangle ABC,  $\overrightarrow{AB}=7\mathbf{i}+\mathbf{j}-\mathbf{k}$ and  $\overrightarrow{AC}=-2\mathbf{i}+5\mathbf{k}$.

Show that $\angle BAC=119.6°$ to 1 d.p.

Hence find the area of triangle ABC, giving your area to 3 s.f.

Point R has position vector$\mathbf{i}+6\mathbf{j}-2\mathbf{k}$and point S has position vector $10\mathbf{i}+13\mathbf{k}$.

Find the unit vector in the direction of $\overrightarrow{RS}$.

Find the angle $\overrightarrow{RS}$ makes with the negative z-axis.

The vector $\overrightarrow{TU}=-12\mathbf{i}+8\mathbf{j}-20\mathbf{k}$.

Explain, giving a reason for your answer, whether the vectors $\overrightarrow{RS}$ and$\overrightarrow{TU}$ are parallel.

The triangle PQR has vertices $P\left(2,-3,1\right)$, $Q\left(-1,-4,3\right)$and $R\left(k,0,3\right)$.  Given that PQR is isosceles, and that $k>1$, find the value of k.

Vectors a and b are defined by 

$$\begin{gathered} \mathbf{a}=\left(p+1\right)\mathbf{i}-7\mathbf{j}+\left(q-3p\right)\mathbf{k} \\ \mathbf{b}=5\mathbf{i}+\left(14q+r\right)\mathbf{j}+\left(1-2r\right)\mathbf{k} \end{gathered}$$  

Given that $\mathbf{a}=\mathbf{b}$, find the values of p, q and r.

A particle of mass 0.5 kg is acted upon by a force of$\left(12\mathbf{i}-4\mathbf{j}+p\mathbf{k}\right)\mathrm{N}$. 

Given that the magnitude of the acceleration experienced by the particle under the influence of this force is 26 m s^-2, find the possible values of p.

An additional second force with a magnitude of qk N now acts on the particle, where  $q<0$.   Under the influence of the resultant of the two forces, the particle now experiences an acceleration of magnitude $8\sqrt{10}\mathrm{m}{\mathrm{s}}^{-2}$.

Explain why this additional information shows that $p>0$.

Hence find the value of q.

Three forces act upon a particle with a mass of 5 kg:

$\begin{array}{rcl}{\mathbf{F}}_1& =& \left(r\mathbf{i}+5\mathbf{j}+\left(r-p\right)\mathbf{k}\right)\mathrm{N}\\ {\mathbf{F}}_2& =& \left(\left(p+q\right)\mathbf{i}-3p\mathbf{j}+7\mathbf{k}\right)\mathrm{N}\\ {\mathbf{F}}_3& =& \left(-\mathbf{i}+r\mathbf{j}-2q\mathbf{k}\right)\mathrm{N}\end{array}$

Under the action of those three forces, the particle is in equilibrium.

Find the values of p, q and r.

A new force, F₄, is added.  Under the combined action of the four forces, the particle experiences an acceleration with a magnitude of  2.2 m s^-2, in the same direction as the vector$\mathit{i}-\mathbf{j}+3\mathbf{k}$.

Find F₄, giving your answer in the form $x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ where x, y and z are given as exact values.  Be sure to include the correct units in your answer.

The diagram below shows a cube whose vertices are O, A, B, C, D, E, F and G.

a, b and c are the vectors $\overrightarrow{OA},\overrightarrow{OB}$ and$\overrightarrow{OC}$respectively.

Using vector methods, prove that the diagonals CD and BF bisect each other.

The coordinates of points A and B are$\left(5,0,-1\right)$and $\left(k,-2,3k\right)$ respectively. Given that the distance from A to B is $6\sqrt{k}$ units, and that point B lies at a distance of $\sqrt{14}$ units from the origin, find the value of k.

In the parallelogram ABCD, AB is parallel to CD, and BC is parallel to AD.

Given that $\overrightarrow{AB}=3\mathbf{i}-\mathbf{j}-2\mathbf{k}$  and  $\overrightarrow{AD}=7\mathbf{i}-\mathbf{j}+4\mathbf{k}$,  find the area of the parallelogram.  Give your answer correct to 3 s.f.

The vector $\overrightarrow{RS}=x\mathbf{i}-9\mathbf{j}+3\mathbf{k}$ makes an angle with the positive x-axis.

Show that $x^2=\frac{90}{{\tan }^2\theta }$.

Given that $\theta$ is acute and that $\cos \theta =\frac{4}{5}$ , find a unit vector in the direction of $\overrightarrow{RS}$.

A, B, C and D are the vertices of a regular tetrahedron.

The coordinates of points $A,B,C$ and D are $\left(1,1,1\right)$, $\left(-8,10,1\right)$, $\left(-3,6,-10\right)$ and $\left(-2k,13,k\right)$ respectively.  Find the coordinates of point $D$.

Vectors a and b are defined by 

$\begin{array}{rcl}\mathbf{a}& =& pq\mathbf{i}-24\mathbf{j}+9r\mathbf{k}\\ \mathbf{b}& =& 6\mathbf{i}+p-r\mathbf{j}+p-3q\mathbf{k}\end{array}$

Given that $\mathbf{a}=3\mathbf{b}$, and that $r>0$, find the values of p, q and r.

A particle of mass 0.2 kg is acted upon by a force of $\left(-3\mathbf{i}+p\mathbf{j}+4\mathbf{k}\right)\mathrm{N}$. 

Given that the magnitude of the acceleration experienced by the particle under the influence of this force is 65 m s-², find the possible values of p.

An additional second force with a magnitude of qj N now acts on the particle, where  $q>0$.   Under the influence of the resultant of the two forces, the particle now experiences an acceleration of magnitude $\frac{5\sqrt{221}}{2}\mathrm{m}{\mathrm{s}}^{-2}$.

Find the possible values of q.

For this question, the unit vectors i and j point in the directions east and north respectively, while the unit vector k points vertically upwards.

A robotic submarine has a mass of 750 kg and is originally moving in a level direction such that the k component of its velocity is zero. In addition to gravity, the forces acting on the submarine are the combined thrust and lift T from its propeller and manoeuvring planes, its buoyancy B, and the water resistance W.  These forces, in newtons, are given by:

$\begin{array}{rcl}\mathbf{T}& =& 600\mathbf{i}-750\mathbf{j}-120\mathbf{k}\\ \mathbf{B}& =& 7360\mathbf{k}\\ \mathbf{W}& =& -500\mathbf{i}+600\mathbf{j}+50\mathbf{k}\end{array}$

Taking the acceleration due to gravity to be $g=9.8$ms^-2, find the magnitude of the acceleration of the submarine.  Give your answer accurate to 3 s.f.

Determine whether the submarine is rising or sinking in the water and determine the angle its acceleration makes with the vector k.  Give your answer accurate to 1 d.p.

The diagram below shows a cuboid whose vertices are O, A, B, C, D, E, F and G.

P is a point on the diagonal OE that divides OE in the ratio $a:b$, where $a>b$.

Show that if line segment BP is extended it will intersect FE, and show that FE is divided in the ratio $\left(a-b\right):b$ by the point of intersection.