SPM KSSM Diagnostic Drill
Form 4 Chapter 8: Vectors (Vektor)
Total Marks: 40 Marks • Time Allowed: 50 Minutes
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A Bahagian A: SPM Paper 1 Format [16 Marks]
Answer all questions
Soalan 1
[4 Marks]
The vectors $\mathbf{u}$ and $\mathbf{v}$ are given by $\mathbf{u} = 6\mathbf{i} - 8\mathbf{j}$ and $\mathbf{v} = k\mathbf{i} + 4\mathbf{j}$, where $k$ is a constant.
(a) Find the unit vector in the direction of $\mathbf{u}$. [2 marks]
(b) If $\mathbf{u}$ is parallel to $\mathbf{v}$, find the value of $k$. [2 marks]
(a) $|\mathbf{u}| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = 10$
$\hat{\mathbf{u}} = \frac{6\mathbf{i} - 8\mathbf{j}}{10} = \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}$
[K1] magnitude 10$\hat{\mathbf{u}} = \frac{6\mathbf{i} - 8\mathbf{j}}{10} = \frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}$
[N1] $\frac{3}{5}\mathbf{i} - \frac{4}{5}\mathbf{j}$
(b) Since $\mathbf{u} \parallel \mathbf{v} \implies \mathbf{u} = \lambda \mathbf{v}$
$6\mathbf{i} - 8\mathbf{j} = \lambda(k\mathbf{i} + 4\mathbf{j}) \implies -8 = 4\lambda \implies \lambda = -2$
$6 = \lambda k \implies 6 = -2k \implies k = -3$.
[K1] equate components$6\mathbf{i} - 8\mathbf{j} = \lambda(k\mathbf{i} + 4\mathbf{j}) \implies -8 = 4\lambda \implies \lambda = -2$
$6 = \lambda k \implies 6 = -2k \implies k = -3$.
[N1] $k = -3$
Soalan 2
[4 Marks]
The position vectors of points $A, B$ and $C$ relative to an origin $O$ are $\vec{OA} = \mathbf{i} + 2\mathbf{j}$, $\vec{OB} = 4\mathbf{i} + 8\mathbf{j}$ and $\vec{OC} = 10\mathbf{i} + 20\mathbf{j}$.
(a) Express $\vec{AB}$ and $\vec{BC}$ in terms of $\mathbf{i}$ and $\mathbf{j}$. [2 marks]
(b) Hence, show that points $A, B$ and $C$ are collinear. [2 marks]
(a) $\vec{AB} = \vec{OB} - \vec{OA} = (4\mathbf{i} + 8\mathbf{j}) - (\mathbf{i} + 2\mathbf{j}) = 3\mathbf{i} + 6\mathbf{j}$
$\vec{BC} = \vec{OC} - \vec{OB} = (10\mathbf{i} + 20\mathbf{j}) - (4\mathbf{i} + 8\mathbf{j}) = 6\mathbf{i} + 12\mathbf{j}$
[N1] $\vec{AB}$$\vec{BC} = \vec{OC} - \vec{OB} = (10\mathbf{i} + 20\mathbf{j}) - (4\mathbf{i} + 8\mathbf{j}) = 6\mathbf{i} + 12\mathbf{j}$
[N1] $\vec{BC}$
(b) Notice that $\vec{BC} = 2(3\mathbf{i} + 6\mathbf{j}) = 2\vec{AB}$. [K1]
Since $\vec{BC} = 2\vec{AB}$ and point $B$ is a common point to both vectors, points $A, B$ and $C$ are collinear.
[N1 explanation]
Soalan 3
[4 Marks]
Given that $(3h - 2)\mathbf{a} + 5\mathbf{b} = (h + 4)\mathbf{a} + (2k - 1)\mathbf{b}$, where $\mathbf{a}$ and $\mathbf{b}$ are non-zero, non-parallel vectors. Find the values of $h$ and $k$.
Since $\mathbf{a} \nparallel \mathbf{b}$, equate the corresponding scalar coefficients:
Coefficients of $\mathbf{a}$: $3h - 2 = h + 4 \implies 2h = 6 \implies h = 3$
[K1] equate $\mathbf{a}$[N1] $h = 3$
Coefficients of $\mathbf{b}$: $5 = 2k - 1 \implies 2k = 6 \implies k = 3$
[K1] equate $\mathbf{b}$[N1] $k = 3$
Soalan 4
[4 Marks]
Two forces $\mathbf{F}_1 = (5\mathbf{i} + 7\mathbf{j})\text{ N}$ and $\mathbf{F}_2 = (7\mathbf{i} - 2\mathbf{j})\text{ N}$ act simultaneously on a particle of mass $2.5\text{ kg}$.
(a) Find the resultant force vector $\mathbf{F}_{\text{net}}$ and its magnitude. [2 marks]
(b) Calculate the acceleration of the particle in $\text{ms}^{-2}$. [2 marks]
(a) $\mathbf{F}_{\text{net}} = \mathbf{F}_1 + \mathbf{F}_2 = (5+7)\mathbf{i} + (7-2)\mathbf{j} = 12\mathbf{i} + 5\mathbf{j}\text{ N}$
$|\mathbf{F}_{\text{net}}| = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = 13\text{ N}$.
[N1] $12\mathbf{i}+5\mathbf{j}$$|\mathbf{F}_{\text{net}}| = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = 13\text{ N}$.
[N1] $13\text{ N}$
(b) Newton's second law: $\mathbf{a} = \frac{\mathbf{F}}{m} = \frac{13\text{ N}}{2.5\text{ kg}} = 5.2\text{ ms}^{-2}$
[K1] $\frac{13}{2.5}$[N1] $5.2\text{ ms}^{-2}$
B Bahagian B: SPM Paper 2 Format [24 Marks]
Detailed solutions required
Soalan 5
[10 Marks]
In quadrilateral $OABC$, $\vec{OA} = \mathbf{a}$ and $\vec{OC} = \mathbf{c}$. Point $D$ lies on $AC$ such that $AD : DC = 1 : 2$. Point $E$ lies on $OB$ such that $\vec{OE} = \frac{3}{4}\vec{OB}$. It is given that $\vec{CB} = 2\mathbf{a}$.
(a) Express $\vec{AC}$ in terms of $\mathbf{a}$ and $\mathbf{c}$. [1 mark]
(b) Express $\vec{OD}$ in terms of $\mathbf{a}$ and $\mathbf{c}$. [2 marks]
(c) Express $\vec{OB}$ and $\vec{OE}$ in terms of $\mathbf{a}$ and $\mathbf{c}$. [3 marks]
(d) Given that $\vec{OF} = h \vec{OD}$ and $\vec{EF} = k(\mathbf{a} - \mathbf{c})$, where $h$ and $k$ are constants. Find the value of $h$ and of $k$. [4 marks]
(a) $\vec{AC} = \vec{OC} - \vec{OA} = \mathbf{c} - \mathbf{a}$
[N1] $\mathbf{c} - \mathbf{a}$
(b) $\vec{AD} = \frac{1}{3}\vec{AC} = \frac{1}{3}(\mathbf{c} - \mathbf{a})$
$\vec{OD} = \vec{OA} + \vec{AD} = \mathbf{a} + \frac{1}{3}(\mathbf{c} - \mathbf{a}) = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{c}$
[K1] path$\vec{OD} = \vec{OA} + \vec{AD} = \mathbf{a} + \frac{1}{3}(\mathbf{c} - \mathbf{a}) = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{c}$
[N1] $\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{c}$
(c) $\vec{OB} = \vec{OC} + \vec{CB} = \mathbf{c} + 2\mathbf{a} = 2\mathbf{a} + \mathbf{c}$ [K1]
$\vec{OE} = \frac{3}{4}\vec{OB} = \frac{3}{4}(2\mathbf{a} + \mathbf{c}) = \frac{3}{2}\mathbf{a} + \frac{3}{4}\mathbf{c}$ [N1]
$\vec{OE} = \frac{3}{4}\vec{OB} = \frac{3}{4}(2\mathbf{a} + \mathbf{c}) = \frac{3}{2}\mathbf{a} + \frac{3}{4}\mathbf{c}$ [N1]
(d) Path: $\vec{OF} = \vec{OE} + \vec{EF} = \left(\frac{3}{2}\mathbf{a} + \frac{3}{4}\mathbf{c}\right) + k(\mathbf{a} - \mathbf{c}) = \left(\frac{3}{2} + k\right)\mathbf{a} + \left(\frac{3}{4} - k\right)\mathbf{c}$ [K1]
Also $\vec{OF} = h\vec{OD} = h\left(\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{c}\right) = \frac{2}{3}h\mathbf{a} + \frac{1}{3}h\mathbf{c}$
Equate $\mathbf{a}$: $\frac{2}{3}h = \frac{3}{2} + k \implies k = \frac{2}{3}h - \frac{3}{2} \quad \text{--- (1)}$
Equate $\mathbf{c}$: $\frac{1}{3}h = \frac{3}{4} - k \implies k = \frac{3}{4} - \frac{1}{3}h \quad \text{--- (2)}$
Equating $k$: $\frac{2}{3}h - \frac{3}{2} = \frac{3}{4} - \frac{1}{3}h \implies h = \frac{3}{4} + \frac{3}{2} = \frac{9}{4} = 2.25$. [K1, N1]
$k = \frac{3}{4} - \frac{1}{3}\left(\frac{9}{4}\right) = \frac{3}{4} - \frac{3}{4} = 0$. [N1]
Also $\vec{OF} = h\vec{OD} = h\left(\frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{c}\right) = \frac{2}{3}h\mathbf{a} + \frac{1}{3}h\mathbf{c}$
Equate $\mathbf{a}$: $\frac{2}{3}h = \frac{3}{2} + k \implies k = \frac{2}{3}h - \frac{3}{2} \quad \text{--- (1)}$
Equate $\mathbf{c}$: $\frac{1}{3}h = \frac{3}{4} - k \implies k = \frac{3}{4} - \frac{1}{3}h \quad \text{--- (2)}$
Equating $k$: $\frac{2}{3}h - \frac{3}{2} = \frac{3}{4} - \frac{1}{3}h \implies h = \frac{3}{4} + \frac{3}{2} = \frac{9}{4} = 2.25$. [K1, N1]
$k = \frac{3}{4} - \frac{1}{3}\left(\frac{9}{4}\right) = \frac{3}{4} - \frac{3}{4} = 0$. [N1]
Soalan 6 • SPM Paper 2 & KBAT
[14 Marks]
Maritime Search & Rescue (SAR)
Straits of Malacca Distress Vessel Vector Interception
Scenario: At 06:00, a disabled fishing vessel drifts from coordinates $(10\mathbf{i} + 30\mathbf{j})\text{ km}$ relative to an offshore radar station at the origin $O$. The ocean current drifts the disabled vessel with a constant velocity of $\mathbf{v}_{\text{drift}} = (4\mathbf{i} - 2\mathbf{j})\text{ km/h}$.
• Let $t$ be the time in hours after 06:00.
(a) Find the position vector $\mathbf{r}_{\text{vessel}}(t)$ of the disabled vessel at any time $t$. [2 marks]
(b) At 07:30 ($t = 1.5$), determine the distance of the vessel from the radar station $O$. [3 marks]
• Let $t$ be the time in hours after 06:00.
(a) Find the position vector $\mathbf{r}_{\text{vessel}}(t)$ of the disabled vessel at any time $t$. [2 marks]
(b) At 07:30 ($t = 1.5$), determine the distance of the vessel from the radar station $O$. [3 marks]
SAR Interception Mission:
At 07:00 ($t = 1.0$), a Malaysian Maritime Enforcement Agency (MMEA) fast interceptor boat departs from the harbor base located at $(-5\mathbf{i} + 5\mathbf{j})\text{ km}$.
• The interceptor travels at a constant velocity $\mathbf{v}_{\text{boat}} = (u\mathbf{i} + v\mathbf{j})\text{ km/h}$.
• The interceptor successfully reaches the disabled vessel at 09:00 ($t = 3.0$).
(c) Find the position vector of the rendezvous point where the rescue occurs. [2 marks]
(d) Calculate the velocity vector $\mathbf{v}_{\text{boat}}$ of the interceptor boat. [4 marks]
(e) Calculate the required cruising speed of the interceptor boat in $\text{km/h}$. [3 marks]
• The interceptor travels at a constant velocity $\mathbf{v}_{\text{boat}} = (u\mathbf{i} + v\mathbf{j})\text{ km/h}$.
• The interceptor successfully reaches the disabled vessel at 09:00 ($t = 3.0$).
(c) Find the position vector of the rendezvous point where the rescue occurs. [2 marks]
(d) Calculate the velocity vector $\mathbf{v}_{\text{boat}}$ of the interceptor boat. [4 marks]
(e) Calculate the required cruising speed of the interceptor boat in $\text{km/h}$. [3 marks]
(a) Position vector $\mathbf{r}_{\text{vessel}}(t)$:
$\mathbf{r}_{\text{vessel}}(t) = \mathbf{r}_0 + t \mathbf{v}_{\text{drift}} = (10\mathbf{i} + 30\mathbf{j}) + t(4\mathbf{i} - 2\mathbf{j}) = (10 + 4t)\mathbf{i} + (30 - 2t)\mathbf{j}\text{ km}$. [K1] $\mathbf{r}_0 + t\mathbf{v}$
[N1] $(10+4t)\mathbf{i}+(30-2t)\mathbf{j}$
$\mathbf{r}_{\text{vessel}}(t) = \mathbf{r}_0 + t \mathbf{v}_{\text{drift}} = (10\mathbf{i} + 30\mathbf{j}) + t(4\mathbf{i} - 2\mathbf{j}) = (10 + 4t)\mathbf{i} + (30 - 2t)\mathbf{j}\text{ km}$. [K1] $\mathbf{r}_0 + t\mathbf{v}$
[N1] $(10+4t)\mathbf{i}+(30-2t)\mathbf{j}$
(b) Distance at $t = 1.5$:
$\mathbf{r}(1.5) = (10 + 4(1.5))\mathbf{i} + (30 - 2(1.5))\mathbf{j} = 16\mathbf{i} + 27\mathbf{j}\text{ km}$. [K1]
Distance $= |\mathbf{r}(1.5)| = \sqrt{16^2 + 27^2} = \sqrt{256 + 729} = \sqrt{985} \approx 31.38\text{ km}$. [K1 Pythagoras]
[N1] $31.38\text{ km}$
$\mathbf{r}(1.5) = (10 + 4(1.5))\mathbf{i} + (30 - 2(1.5))\mathbf{j} = 16\mathbf{i} + 27\mathbf{j}\text{ km}$. [K1]
Distance $= |\mathbf{r}(1.5)| = \sqrt{16^2 + 27^2} = \sqrt{256 + 729} = \sqrt{985} \approx 31.38\text{ km}$. [K1 Pythagoras]
[N1] $31.38\text{ km}$
(c) Rendezvous at 09:00 ($t = 3.0$):
$\mathbf{r}_{\text{rendezvous}} = \mathbf{r}(3) = (10 + 4(3))\mathbf{i} + (30 - 2(3))\mathbf{j} = 22\mathbf{i} + 24\mathbf{j}\text{ km}$. [K1] $t=3$
[N1] $22\mathbf{i} + 24\mathbf{j}\text{ km}$
$\mathbf{r}_{\text{rendezvous}} = \mathbf{r}(3) = (10 + 4(3))\mathbf{i} + (30 - 2(3))\mathbf{j} = 22\mathbf{i} + 24\mathbf{j}\text{ km}$. [K1] $t=3$
[N1] $22\mathbf{i} + 24\mathbf{j}\text{ km}$
(d) Interceptor velocity:
Boat departs at $t = 1.0$ (07:00) and arrives at $t = 3.0$ (09:00) $\implies \Delta t = 2.0\text{ hours}$.
Initial position of boat: $\mathbf{r}_{\text{base}} = -5\mathbf{i} + 5\mathbf{j}$.
Displacement vector $\vec{D} = \mathbf{r}_{\text{rendezvous}} - \mathbf{r}_{\text{base}} = (22\mathbf{i} + 24\mathbf{j}) - (-5\mathbf{i} + 5\mathbf{j}) = 27\mathbf{i} + 19\mathbf{j}\text{ km}$. [K1]
$\mathbf{v}_{\text{boat}} = \frac{\vec{D}}{\Delta t} = \frac{27\mathbf{i} + 19\mathbf{j}}{2.0} = 13.5\mathbf{i} + 9.5\mathbf{j}\text{ km/h}$. [K1 displacement]
[N1] $13.5\mathbf{i} + 9.5\mathbf{j}\text{ km/h}$
Boat departs at $t = 1.0$ (07:00) and arrives at $t = 3.0$ (09:00) $\implies \Delta t = 2.0\text{ hours}$.
Initial position of boat: $\mathbf{r}_{\text{base}} = -5\mathbf{i} + 5\mathbf{j}$.
Displacement vector $\vec{D} = \mathbf{r}_{\text{rendezvous}} - \mathbf{r}_{\text{base}} = (22\mathbf{i} + 24\mathbf{j}) - (-5\mathbf{i} + 5\mathbf{j}) = 27\mathbf{i} + 19\mathbf{j}\text{ km}$. [K1]
$\mathbf{v}_{\text{boat}} = \frac{\vec{D}}{\Delta t} = \frac{27\mathbf{i} + 19\mathbf{j}}{2.0} = 13.5\mathbf{i} + 9.5\mathbf{j}\text{ km/h}$. [K1 displacement]
[N1] $13.5\mathbf{i} + 9.5\mathbf{j}\text{ km/h}$
(e) Required Cruising Speed:
Speed $= |\mathbf{v}_{\text{boat}}| = \sqrt{(13.5)^2 + (9.5)^2} = \sqrt{182.25 + 90.25} = \sqrt{272.5} \approx 16.51\text{ km/h}$ (or $\approx 8.91\text{ knots}$). [K1 speed magnitude]
[N1] $16.51\text{ km/h}$
Speed $= |\mathbf{v}_{\text{boat}}| = \sqrt{(13.5)^2 + (9.5)^2} = \sqrt{182.25 + 90.25} = \sqrt{272.5} \approx 16.51\text{ km/h}$ (or $\approx 8.91\text{ knots}$). [K1 speed magnitude]
[N1] $16.51\text{ km/h}$