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E Math (IGCSE)

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

IB AI HL Math

IB AI SL Math

IP Math

N(A) A Math

N(A) E Math

PSLE

PSLE Foundation Mathematics

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data-width="420" data-height="360" data-proportional style="width:420px;height:360px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:420px;--graph-aspect:420 / 360;"></div>
<p>An oblique pyramid has a plane base $ABCD$ in the shape of a parallelogram. The coordinates of $A$, $B$ and $C$ are $(5,-4,1)$, $(5,4,0)$ and $(-5,4,2)$ respectively. The apex of the pyramid is at $E(0,0,10)$ (see diagram).</p>
<ol class="i-list"><li data-list-item-id="imported-item-block-2-item-0" data-list-item-path="imported-item-block-2-item-0"><div class="list-item-paragraph">Find the coordinates of $D$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="imported-item-block-2-item-1" data-list-item-path="imported-item-block-2-item-1"><div class="list-item-paragraph">Find the cartesian equation of face $BCE$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="imported-item-block-2-item-2" data-list-item-path="imported-item-block-2-item-2"><div class="list-item-paragraph">Find the angle between face $BCE$ and the base of the pyramid.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="imported-item-block-2-item-3" data-list-item-path="imported-item-block-2-item-3"><div class="list-item-paragraph">Find the shortest distance from the midpoint of edge $AD$ to face $BCE$.<span class="inline-marks">[5]</span></div></li></ol>

An oblique pyramid has a plane base $ABCD$ in the shape of a parallelogram. The coordinates of $A$, $B$ and $C$ are $(5,-4,1)$, $(5,4,0)$ and $(-5,4,2)$ respectively. The apex of the pyramid is at $E(0,0,10)$ (see diagram).

Find the cartesian equation of face $BCE$.

Find the angle between face $BCE$ and the base of the pyramid.

Find the shortest distance from the midpoint of edge $AD$ to face $BCE$.

Distance from a Point to a Plane

Equations of Planes

Angle Between Two Planes

Vectors in a Parallelogram

Ratio Theorem and Section Formula

N2018 P2 Q3

Vectors

<ol class="abc-list"><li data-list-item-id="h2-2025-yijc-promo-q8-yijc-q8-item-block-0-item-0" data-list-item-path="h2-2025-yijc-promo-q8-yijc-q8-item-block-0-item-0"><div class="list-item-paragraph">With reference to the origin $O$, points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. Point $P$ lies on $AB$, such that&nbsp;$AP:PB=2:3$. Given that&nbsp;$OP$ and $AB$ are perpendicular, show that&nbsp;$\mathbf{a}\cdot\mathbf{b}=3|\mathbf{a}|^2-2|\mathbf{b}|^2$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-yijc-promo-q8-yijc-q8-item-block-0-item-1" data-list-item-path="h2-2025-yijc-promo-q8-yijc-q8-item-block-0-item-1"><div class="list-item-paragraph">The plane $\pi$ has equation $\mathbf{r}=\begin{pmatrix}0\\2\\1\end{pmatrix}+\lambda\begin{pmatrix}-1\\2\\3\end{pmatrix}+\mu\begin{pmatrix}3\\0\\-1\end{pmatrix},\ \lambda,\mu\in\mathbb{R}.$</div><ol class="i-list"><li data-list-item-id="item-1789031430179-n1av0zoyd" data-list-item-path="item-1789031430179-n1av0zoyd"><div class="list-item-paragraph"><p>Show that a cartesian equation of $\pi$ is $x-4y+3z=-5$.</p><span class="inline-marks">[3]</span></div></li></ol><div class="list-item-paragraph"><p>The line $l_1$ has equation $\mathbf{r}=\begin{pmatrix}6\\-1\\4\end{pmatrix}+t\begin{pmatrix}1\\1\\1\end{pmatrix},\ t\in\mathbb{R}.$</p></div><ol class="i-list" start="2"><li data-list-item-id="h2-2025-yijc-promo-q8-yijc-q8-item-block-0-item-1-sub-4" data-list-item-path="h2-2025-yijc-promo-q8-yijc-q8-item-block-0-item-1/h2-2025-yijc-promo-q8-yijc-q8-item-block-0-item-1-sub-4"><div class="list-item-paragraph"><p>Find the acute angle between $l_1$ and $\pi$.</p><span class="inline-marks">[3]</span></div></li></ol><div class="list-item-after-content"><p>The line $l_2$ passes through the point $Q(-2,-1,2)$ and is parallel to $2\mathbf{i}+3\mathbf{j}+\mathbf{k}$.</p></div><div class="list-item-after-content"><ol class="i-list" start="3"><li data-list-item-id="item-1789031789556-jh2eio6wg" data-list-item-path="item-1789031789556-jh2eio6wg"><div class="list-item-paragraph"><p>Find the position vector of the foot of the perpendicular from $Q$ to $\pi$.</p><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1789031809165-lneq3t01d" data-list-item-path="item-1789031809165-lneq3t01d"><div class="list-item-paragraph"><p>Find the distance between $l_2$ and $\pi$.</p><span class="inline-marks">[2]</span></div></li></ol></div></li></ol>

With reference to the origin $O$, points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. Point $P$ lies on $AB$, such that $AP:PB=2:3$. Given that $OP$ and $AB$ are perpendicular, show that $\mathbf{a}\cdot\mathbf{b}=3|\mathbf{a}|^2-2|\mathbf{b}|^2$.

Perpendicular Projection and Reflection in a Plane

Angle Between a Line and a Plane

Perpendicular Vectors Using the Scalar Product

2025 YIJC Promo Q8

<p>With reference to the origin $O$, the points $A$, $B$ and $X$ have position vectors $3\mathbf{i}+2\mathbf{j}+\mathbf{k}$, $\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ and $11\mathbf{i}+10\mathbf{j}+9\mathbf{k}$ respectively. The point $C$ lies on $OA$ produced such that&nbsp;$OA:AC=1:\alpha$, where&nbsp;$\alpha$ is a constant. The point $D$ lies on $OB$ produced such that&nbsp;$OB:BD=1:\beta$, where&nbsp;$\beta$ is a constant.</p>
<p>It is given that&nbsp;$OCXD$ is a parallelogram.</p>
<ol class="abc-list"><li data-list-item-id="h2-2025-tjc-promo-q3-tjc-q3-item-block-2-item-0" data-list-item-path="h2-2025-tjc-promo-q3-tjc-q3-item-block-2-item-0"><div class="list-item-paragraph">Write down the position vectors of $C$ and $D$ in terms of $\alpha$ and $\beta$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-2025-tjc-promo-q3-tjc-q3-item-block-2-item-1" data-list-item-path="h2-2025-tjc-promo-q3-tjc-q3-item-block-2-item-1"><div class="list-item-paragraph">Show that&nbsp;$\beta=1$ and find the value of $\alpha$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-tjc-promo-q3-tjc-q3-item-block-2-item-2" data-list-item-path="h2-2025-tjc-promo-q3-tjc-q3-item-block-2-item-2"><div class="list-item-paragraph">Find the exact shortest distance from $X$ to line $OD$.<span class="inline-marks">[3]</span></div></li></ol>

With reference to the origin $O$, the points $A$, $B$ and $X$ have position vectors $3\mathbf{i}+2\mathbf{j}+\mathbf{k}$, $\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ and $11\mathbf{i}+10\mathbf{j}+9\mathbf{k}$ respectively. The point $C$ lies on $OA$ produced such that $OA:AC=1:\alpha$, where $\alpha$ is a constant. The point $D$ lies on $OB$ produced such that $OB:BD=1:\beta$, where $\beta$ is a constant.

It is given that $OCXD$ is a parallelogram.

Write down the position vectors of $C$ and $D$ in terms of $\alpha$ and $\beta$.

Show that $\beta=1$ and find the value of $\alpha$.

Find the exact shortest distance from $X$ to line $OD$.

Distance from a Point to a Line

2025 TJC Promo Q3

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<p>The diagram shows a tetrahedron with vertices $O$, $A$, $B$ and $C$. Points $A$, $B$ and $C$ have position vectors with respect to the origin $O$ given by $\overrightarrow{OA}=\begin{pmatrix}2\\-1\\2\end{pmatrix}$, $\overrightarrow{OB}=\begin{pmatrix}0\\3\\1\end{pmatrix}$ and $\overrightarrow{OC}=\begin{pmatrix}3\\0\\4\end{pmatrix}$. The point $D$ lies on $BC$, between $B$ and $C$, such that&nbsp;$BD:DC=1:2$.</p>
<ol class="abc-list"><li data-list-item-id="h2-2025-nyjc-promo-q9-nyjc-q9-item-block-2-item-0" data-list-item-path="h2-2025-nyjc-promo-q9-nyjc-q9-item-block-2-item-0"><div class="list-item-paragraph">Find the position vector of $D$<span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-2025-nyjc-promo-q9-nyjc-q9-item-block-2-item-1" data-list-item-path="h2-2025-nyjc-promo-q9-nyjc-q9-item-block-2-item-1"><div class="list-item-paragraph">Find a cartesian equation of face $ABC$ and state its exact distance from $O$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="h2-2025-nyjc-promo-q9-nyjc-q9-item-block-2-item-2" data-list-item-path="h2-2025-nyjc-promo-q9-nyjc-q9-item-block-2-item-2"><div class="list-item-paragraph">Hence find the acute angle between $OD$ and face $ABC$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-2025-nyjc-promo-q9-nyjc-q9-item-block-2-item-3" data-list-item-path="h2-2025-nyjc-promo-q9-nyjc-q9-item-block-2-item-3"><div class="list-item-paragraph">Find the coordinates of the point on $OD$ which is closest to $A$.<span class="inline-marks">[3]</span></div></li></ol>

The diagram shows a tetrahedron with vertices $O$, $A$, $B$ and $C$. Points $A$, $B$ and $C$ have position vectors with respect to the origin $O$ given by $\overrightarrow{OA}=\begin{pmatrix}2\\-1\\2\end{pmatrix}$, $\overrightarrow{OB}=\begin{pmatrix}0\\3\\1\end{pmatrix}$ and $\overrightarrow{OC}=\begin{pmatrix}3\\0\\4\end{pmatrix}$. The point $D$ lies on $BC$, between $B$ and $C$, such that $BD:DC=1:2$.

Find a cartesian equation of face $ABC$ and state its exact distance from $O$.

Hence find the acute angle between $OD$ and face $ABC$.

Find the coordinates of the point on $OD$ which is closest to $A$.

Perpendicular Projection and Reflection in a Line

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<p>The diagram shows a unit circle with centre $O$ and diameter $AB$. The point $C$ lies on the circumference of the circle. The position vectors of $A$, $B$ and $C$ are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively.</p>
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<ol class="abc-list"><li data-list-item-id="h2-2025-cjc-promo-q6-cjc-q6-item-block-2-item-0" data-list-item-path="h2-2025-cjc-promo-q6-cjc-q6-item-block-2-item-0"><div class="list-item-paragraph">By considering an appropriate scalar product, show that&nbsp;$AC$ and $BC$ are perpendicular.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="h2-2025-cjc-promo-q6-cjc-q6-item-block-2-item-1" data-list-item-path="h2-2025-cjc-promo-q6-cjc-q6-item-block-2-item-1"><div class="list-item-paragraph">Give a geometrical interpretation of $|\mathbf{a}\cdot(\mathbf{c}-\mathbf{a})|$.<span class="inline-marks">[1]</span></div></li></ol>
<p>The point $D$ is such that&nbsp;$ABCD$ is a parallelogram. The point $M$ is the mid-point of $AD$ and the point $R$ lies on $OA$ produced such that&nbsp;$OA:AR=1:2$.</p>
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<ol class="abc-list" start="3"><li data-list-item-id="h2-2025-cjc-promo-q6-cjc-q6-item-block-5-item-0" data-list-item-path="h2-2025-cjc-promo-q6-cjc-q6-item-block-5-item-0"><div class="list-item-paragraph">By considering $\overrightarrow{CM}\times\overrightarrow{CR}$ or otherwise, show that the points $C$, $M$ and $R$ are collinear.<span class="inline-marks">[3]</span></div></li></ol>

The diagram shows a unit circle with centre $O$ and diameter $AB$. The point $C$ lies on the circumference of the circle. The position vectors of $A$, $B$ and $C$ are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively.

By considering an appropriate scalar product, show that $AC$ and $BC$ are perpendicular.

Give a geometrical interpretation of $|\mathbf{a}\cdot(\mathbf{c}-\mathbf{a})|$.

The point $D$ is such that $ABCD$ is a parallelogram. The point $M$ is the mid-point of $AD$ and the point $R$ lies on $OA$ produced such that $OA:AR=1:2$.

By considering $\overrightarrow{CM}\times\overrightarrow{CR}$ or otherwise, show that the points $C$, $M$ and $R$ are collinear.

Collinearity Using Vectors

Projection Lengths

2025 CJC Promo Q6

<p>The figure shows a right-angled triangle $ABC$ in which $A$, $B$ and $C$ are $(-3,\ 18)$, $(k,\ 0)$ and $(15,\ 4)$ respectively. $AB$ cuts the $y$-axis at $P$. The point $Q$ lies on $AC$ and $PQ$ is parallel to $BC$.</p>
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<ol class="i-list"><li data-list-item-id="cg-q10-i" data-list-item-path="cg-q10-i"><div class="list-item-paragraph">Given that&nbsp;$k&lt;5$, find the value of $k$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="cg-q10-ii" data-list-item-path="cg-q10-ii"><div class="list-item-paragraph">Find the coordinates of $P$ and show that&nbsp;$P$ is the midpoint of $AB$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="cg-q10-iii" data-list-item-path="cg-q10-iii"><div class="list-item-paragraph">Hence find the coordinates of $Q$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="cg-q10-iv" data-list-item-path="cg-q10-iv"><div class="list-item-paragraph">The point $D$ is such that&nbsp;$ABCD$ is a rectangle. Find the coordinates of $D$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="cg-q10-v" data-list-item-path="cg-q10-v"><div class="list-item-paragraph">Find the area of the rectangle $ABCD$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="cg-q10-vi" data-list-item-path="cg-q10-vi"><div class="list-item-paragraph">The point $F$ lies on $BC$ such that&nbsp;$BF : FC = 1 : 3$. Find the coordinates of $F$.<span class="inline-marks">[2]</span></div></li></ol>

The figure shows a right-angled triangle $ABC$ in which $A$, $B$ and $C$ are $(-3,\ 18)$, $(k,\ 0)$ and $(15,\ 4)$ respectively. $AB$ cuts the $y$-axis at $P$. The point $Q$ lies on $AC$ and $PQ$ is parallel to $BC$.

Given that $k<5$, find the value of $k$.

Find the coordinates of $P$ and show that $P$ is the midpoint of $AB$.

The point $D$ is such that $ABCD$ is a rectangle. Find the coordinates of $D$.

The point $F$ lies on $BC$ such that $BF : FC = 1 : 3$. Find the coordinates of $F$.

Area of Rectilinear Figures

Parallel and Perpendicular Lines

Equation of a Straight Line

Midpoint of a Line Segment

Right-angled triangle completed to a rectangle

Coordinate Geometry

<p>The diagram shows a trapezium $ABCD$ in which $A$ is $(1,\ 2)$, $B$ is $(5,\ 10)$ and $D$ is $(9,\ 3)$. Angle $ABC$ is $90^\circ$ and $AB$ is parallel to $DC$.</p>
<div class="graph-block" data-graph-config="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" data-width="395" data-height="250" data-proportional style="width:395px;height:250px;max-width:100%;display:flow-root;margin:0 auto;margin-bottom:0px;--graph-width:395px;--graph-aspect:395 / 250;"></div>
<ol class="i-list"><li data-list-item-id="cg-q09-i" data-list-item-path="cg-q09-i"><div class="list-item-paragraph">Find the equation of $CD$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="cg-q09-ii" data-list-item-path="cg-q09-ii"><div class="list-item-paragraph">Find the equation of $BC$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="cg-q09-iii" data-list-item-path="cg-q09-iii"><div class="list-item-paragraph">Hence find the coordinates of $C$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="cg-q09-iv" data-list-item-path="cg-q09-iv"><div class="list-item-paragraph">Find the area of the trapezium $ABCD$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="cg-q09-v" data-list-item-path="cg-q09-v"><div class="list-item-paragraph">The point $E$ lies on $AC$ such that&nbsp;$AE : EC = 3 : 2$. Find the coordinates of $E$.<span class="inline-marks">[2]</span></div></li></ol>

The diagram shows a trapezium $ABCD$ in which $A$ is $(1,\ 2)$, $B$ is $(5,\ 10)$ and $D$ is $(9,\ 3)$. Angle $ABC$ is $90^\circ$ and $AB$ is parallel to $DC$.

The point $E$ lies on $AC$ such that $AE : EC = 3 : 2$. Find the coordinates of $E$.

Intersection of Two Lines

Right-angled trapezium and a point dividing a diagonal

<p>The position vectors of the points $X$, $Y$ and $Z$, relative to an origin are $3\mathbf{m}$, $2\mathbf{n}$ and $\frac{1}{2}\mathbf{m}-\mathbf{n}$ respectively.</p>
<p>Given that&nbsp;$\overrightarrow{XY} = \lambda \overrightarrow{XP}$ and $\overrightarrow{OZ} = \mu \overrightarrow{OP}$, where&nbsp;$\lambda$ and $\mu$ are non-zero constants,</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">express $\overrightarrow{OP}$ in terms of $m$, $n$ and $\lambda$,</div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">express $\overrightarrow{OP}$ in terms of $m$, $n$ and $\mu$,</div></li><li data-list-item-id="item-iii"><div class="list-item-paragraph">hence, evaluate $\lambda$ and $\mu$.</div></li></ol>

The position vectors of the points $X$, $Y$ and $Z$, relative to an origin are $3\mathbf{m}$, $2\mathbf{n}$ and $\frac{1}{2}\mathbf{m}-\mathbf{n}$ respectively.

Given that $\overrightarrow{XY} = \lambda \overrightarrow{XP}$ and $\overrightarrow{OZ} = \mu \overrightarrow{OP}$, where $\lambda$ and $\mu$ are non-zero constants,

Vector Algebra

<p>The diagram shows a line $DE$ which cuts the $y$-axis at $P$ and a line through $F$ meets the $y$-axis at $Q$ such that angle $FPQ =$ angle $FQP$.</p>
<p>The coordinates of $D$ and $E$ are $(-4, 0)$ and $(10, 7)$ respectively.</p>
<p>$F$ is a point on the line $DE$ such that&nbsp;$DF : FE = 4 : 3$.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1782712311122-rmouexzadr.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI3MTIzMTExMjItcm1vdWV4emFkci5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjk1ODMxLCJleHAiOjE3OTI2OTYxMzF9.WUKWLORyVJpRL1_kM2-GVdAs_xfi9vkH7iljjykPoHw" style="max-width: min(380px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Find the coordinates of $F$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Find the equation of the straight line $FQ$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-3" data-list-item-path="item-3"><div class="list-item-paragraph">Find the area of the triangle $QFE$.<span class="inline-marks">[3]</span></div></li></ol>

The diagram shows a line $DE$ which cuts the $y$-axis at $P$ and a line through $F$ meets the $y$-axis at $Q$ such that angle $FPQ =$ angle $FQP$.

The coordinates of $D$ and $E$ are $(-4, 0)$ and $(10, 7)$ respectively.

$F$ is a point on the line $DE$ such that $DF : FE = 4 : 3$.

2024 CGSS P2 Q5

<p>With respect to the origin $O$, the fixed points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively, where&nbsp;$\mathbf{a}$ and $\mathbf{b}$ are non-zero and non-parallel.</p>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-paragraph">The variable point $R$ has position vector $\mathbf{r} = \lambda \mathbf{a} + (1 - \lambda)\mathbf{b}$, where&nbsp;$\lambda$ is a real parameter. Describe geometrically the set of all possible positions of $R$.<span class="inline-marks">[1]</span></div></li></ol>
<p>It is given that the angle $AOB$ is $90^\circ$.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-b"><div class="list-item-paragraph">Explain why $\mathbf{a} \cdot \mathbf{b} = 0$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-c"><div class="list-item-paragraph">Among the set of all possible points $R$, the point $R^*$ is the closest to the origin $O$. Find the position vector of $R^*$. Hence state the ratio $AR^* : BR^*$ in terms of magnitudes of $\mathbf{a}$ and $\mathbf{b}$.<span class="inline-marks">[4]</span></div></li></ol>

With respect to the origin $O$, the fixed points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively, where $\mathbf{a}$ and $\mathbf{b}$ are non-zero and non-parallel.

It is given that the angle $AOB$ is $90^\circ$.

Geometrical Interpretation of Vector Equations

2024 RI Promo Q2

<p>Relative to the origin $O$, the position vectors of the points $A$ and $B$ are given by $\mathbf{a}$ and $\mathbf{b}$ respectively. The point $P$ is on $OA$ such that&nbsp;$OP = 3PA$ and point $Q$ is on $AB$ such that&nbsp;$AQ = kAB$, where&nbsp;$k$ is a constant and $0 &lt; k &lt; 1$.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">Express $\overrightarrow{OP}$ and $\overrightarrow{OQ}$ in terms of $\mathbf{a}$, $\mathbf{b}$ and $k$, where applicable.<span class="inline-marks">[2]</span></div></li></ol>
<p>Point $R$ is on $PQ$ produced such that&nbsp;$PQ : QR = 4 : 5$.</p>
<ol class="i-list" start="2"><li data-list-item-id="item-ii"><div class="list-item-paragraph">Given that the points $O$, $B$ and $R$ are collinear, find the value of $k$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-iii"><div class="list-item-paragraph">Show that the area of triangle $OPR$ can be written as&nbsp;$s\left| \mathbf{a}\times \mathbf{b} \right|$, where&nbsp;$s$ is a constant to be found.<span class="inline-marks">[2]</span></div></li></ol>

Relative to the origin $O$, the position vectors of the points $A$ and $B$ are given by $\mathbf{a}$ and $\mathbf{b}$ respectively. The point $P$ is on $OA$ such that $OP = 3PA$ and point $Q$ is on $AB$ such that $AQ = kAB$, where $k$ is a constant and $0 < k < 1$.

Point $R$ is on $PQ$ produced such that $PQ : QR = 4 : 5$.

Areas of Triangles and Parallelograms Using Vectors

2024 MI PRU2 P2 Q3

<p>Referred to the origin $O$, the points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$. $C$ lies on $AB$ such that&nbsp;$AC : AB = 1 : 3$.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">Find $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">Show that the area of triangle $OAC$ is given by $k|\mathbf{a} \times \mathbf{b}|$, where&nbsp;$k$ is a constant to be determined.<span class="inline-marks">[2]</span></div></li></ol>
<p>It is now given that&nbsp;$\mathbf{a}$ and $\mathbf{b}$ are unit vectors.</p>
<ol class="i-list" start="3"><li data-list-item-id="item-iii"><div class="list-item-paragraph">Using the result in part <strong>(i)</strong>, find the value of $(2\mathbf{a} - \mathbf{b}) \cdot \mathbf{c}$.<span class="inline-marks">[3]</span></div></li></ol>

Referred to the origin $O$, the points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$. $C$ lies on $AB$ such that $AC : AB = 1 : 3$.

It is now given that $\mathbf{a}$ and $\mathbf{b}$ are unit vectors.

Scalar (Dot) Product

2024 MI PRU1 Q4

<p>Referred to the origin $O$, the points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively such that&nbsp;$|\mathbf{a}|=5$ and $|\mathbf{b}|=3$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">It is given that&nbsp;$|2\mathbf{b} - \mathbf{a}| = 7$. Using a suitable scalar product, show that&nbsp;$\mathbf{a} \cdot \mathbf{b} = 3$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Deduce the length of projection of $\overrightarrow{OA}$ onto $\overrightarrow{OB}$.</span><span class="inline-marks">[1]</span></div></li></ol>
<p>The point $P$ lies on $AB$ such that&nbsp;$AP:PB = 1:k$, where&nbsp;$k$ is a positive constant.</p>
<ol class="abc-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Write down the position vector of $P$ in terms of $k$, $\mathbf{a}$ and $\mathbf{b}$ and show that the area of triangle $OAP$ is $\frac{3\sqrt{6}}{k+1}$.</span><span class="inline-marks">[5]</span></div></li></ol>

Referred to the origin $O$, the points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively such that $|\mathbf{a}|=5$ and $|\mathbf{b}|=3$.

It is given that $|2\mathbf{b} - \mathbf{a}| = 7$. Using a suitable scalar product, show that $\mathbf{a} \cdot \mathbf{b} = 3$.

Deduce the length of projection of $\overrightarrow{OA}$ onto $\overrightarrow{OB}$.

The point $P$ lies on $AB$ such that $AP:PB = 1:k$, where $k$ is a positive constant.

Write down the position vector of $P$ in terms of $k$, $\mathbf{a}$ and $\mathbf{b}$ and show that the area of triangle $OAP$ is $\frac{3\sqrt{6}}{k+1}$.

<p>In triangle $AOB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $C$ lies on $AB$ such that&nbsp;$AC:AB = 1:3$.</p>
<p>Express $\overrightarrow{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$, giving your answer in the simplest form.</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1776508414841-i8f3zjzuriq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzY1MDg0MTQ4NDEtaThmM3pqenVyaXEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY5NTgzMSwiZXhwIjoxNzkyNjk2MTMxfQ.NLmgnf5mhvSPglUvE2O6m531c05OX2AekqsAk_NbirQ" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In triangle $AOB$, $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$. $C$ lies on $AB$ such that $AC:AB = 1:3$.

Express $\overrightarrow{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$, giving your answer in the simplest form.

<p>Referred to the origin $O$, points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. Point $C$ lies on $OA$, between $O$ and $A$, such that&nbsp;$OC : CA = 3 : 2$. Point $D$ lies on $OB$, between $O$ and $B$, such that&nbsp;$OD : DB = 5 : 6$.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the position vectors $\overrightarrow{OC}$ and $\overrightarrow{OD}$, giving your answers in terms of $\mathbf{a}$ and $\mathbf{b}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Show that the vector equation of the line $BC$ can be written as&nbsp;$\mathbf{r} = \frac{3}{5} \lambda \mathbf{a} + (1 - \lambda) \mathbf{b}$, where&nbsp;$\lambda$ is a parameter. Find in similar form the vector equation of the line $AD$ in terms of a parameter $\mu$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find, in terms of $\mathbf{a}$ and $\mathbf{b}$, the position vector of the point $E$ where the lines $BC$ and $AD$ meet and find the ratio $AE : ED$.<span class="inline-marks">[5]</span></div></li></ol>

Referred to the origin $O$, points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. Point $C$ lies on $OA$, between $O$ and $A$, such that $OC : CA = 3 : 2$. Point $D$ lies on $OB$, between $O$ and $B$, such that $OD : DB = 5 : 6$.

Show that the vector equation of the line $BC$ can be written as $\mathbf{r} = \frac{3}{5} \lambda \mathbf{a} + (1 - \lambda) \mathbf{b}$, where $\lambda$ is a parameter. Find in similar form the vector equation of the line $AD$ in terms of a parameter $\mu$.

Find, in terms of $\mathbf{a}$ and $\mathbf{b}$, the position vector of the point $E$ where the lines $BC$ and $AD$ meet and find the ratio $AE : ED$.

Equations of Lines

N2015 P1 Q7

<p>With respect to an origin $O$, the points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively, where&nbsp;$\mathbf{a}$ and $\mathbf{b}$ are not parallel.</p>
<ol class="i-list"><li><div class="list-item-paragraph">It is given that&nbsp;$B$ lies on the line segment $AC$, such that&nbsp;$\overrightarrow{BC}=2\mathbf{b}-k\mathbf{a}$. State the value of the constant $k$. Hence, 	find $\overrightarrow{OC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">$\overrightarrow{ON}$ is the projection vectors of $\overrightarrow{OB}$ onto $\overrightarrow{OA}$, where&nbsp;$\overrightarrow{ON}=\frac{3}{4}\mathbf{a}$ and $\mathbf{a}\cdot \mathbf{b}=4$. Show that the length of $OA$ is $\frac{4\sqrt{3}}{3}$.	<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Given also that&nbsp;$\mathbf{b}=\left( \begin{matrix}
   2  \\
   -2  \\
   2  \\
\end{matrix} \right)$, find the angle between $\overrightarrow{OA}$ and  $\overrightarrow{OB}$.	<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Given that&nbsp;$OADC$ is a parallelogram and $E$ lies on line segment $AC$ such that&nbsp;$AE:AC=2:3$, find $\overrightarrow{DE}$.<span class="inline-marks">[3]</span></div></li></ol>

With respect to an origin $O$, the points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively, where $\mathbf{a}$ and $\mathbf{b}$ are not parallel.

Vector Projections

Angles Between Vectors

Magnitude of a Vector

<p>With respect to the origin $O$ , the position vectors of two points $A$ and $B$  are given by $\mathbf{a}$ and $\mathbf{b}$ respectively. $C$ on $AB$ produced such that&nbsp;$AC=2AB$. Find the position vector of $C$ in terms of $\mathbf{a}$ and $\mathbf{b}$.</p>

With respect to the origin $O$ , the position vectors of two points $A$ and $B$  are given by $\mathbf{a}$ and $\mathbf{b}$ respectively. $C$ on $AB$ produced such that $AC=2AB$. Find the position vector of $C$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

<p>With respect to the origin $O$ , the position vectors of two points $A$ and $B$  are $\left( \mathbf{i}+\mathbf{j}-2\mathbf{k} \right)$ and $\left( -2\mathbf{i}+7\mathbf{j}-\mathbf{k} \right)$ respectively. $C$ is a point produced on line $AB$ such that&nbsp;$5AB=2AC$. Find the position vector of $C$.</p>

With respect to the origin $O$ , the position vectors of two points $A$ and $B$  are $\left( \mathbf{i}+\mathbf{j}-2\mathbf{k} \right)$ and $\left( -2\mathbf{i}+7\mathbf{j}-\mathbf{k} \right)$ respectively. $C$ is a point produced on line $AB$ such that $5AB=2AC$. Find the position vector of $C$.

<p>Referred to the origin $O$, points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. Point $C$ lies on $OA$, such that&nbsp;$OC:CA=2:1$. Point $D$ lies on $OB$, such that&nbsp;$OD:DB=\lambda:\mu$. It is given that the area of triangle $ABD$ is half the area of triangle $ABC$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Show the area of triangle $ABD$ is given by $\frac{\mu}{2(\lambda+\mu)} |\mathbf{a} \times \mathbf{b}|$. Hence find the ratio $\lambda:\mu$.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">The point $E$ has position vector $\frac{1}{4}\mathbf{a}+\frac{5}{8}\mathbf{b}$. Show that&nbsp;$A$, $E$ and $D$ are collinear.</span><span class="inline-marks">[3]</span></div></li></ol>
<p>It is further given that the angle $AOB$ is $\frac{\pi}{4}$ and $O$ lies on the perpendicular bisector of the line segment $AB$.</p>
<ol class="abc-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Find the length of projection of $\mathbf{a}$ on $\mathbf{b}$, giving your answer in terms of $|\mathbf{b}|$. Hence find the position vector of the point $F$, the foot of perpendicular from $A$ to $OB$.</span><span class="inline-marks">[3]</span></div></li></ol>

Referred to the origin $O$, points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. Point $C$ lies on $OA$, such that 
$OC:CA=2:1$. Point $D$ lies on $OB$, such that $OD:DB=\lambda:\mu$. It is given that the area of triangle $ABD$ is half the area of triangle $ABC$.

The point $E$ has position vector $\frac{1}{4}\mathbf{a}+\frac{5}{8}\mathbf{b}$. Show that $A$, $E$ and $D$ are collinear.

It is further given that the angle $AOB$ is $\frac{\pi}{4}$ and $O$ lies on the perpendicular bisector of the line segment $AB$.

Find the length of projection of $\mathbf{a}$ on $\mathbf{b}$, giving your answer in terms of $|\mathbf{b}|$. Hence find the position vector of the point $F$, the foot of perpendicular from $A$ to $OB$.

2025 VJC P1 Q3

<p>For any vectors $\mathbf{m}$ and $\mathbf{n}$, explain why $\mathbf{m} \cdot (\mathbf{m} \times \mathbf{n}) = 0$.<span class="inline-marks">[1]</span></p>
<p>With respect to the origin $O$, the points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively. $O$, $A$, $B$ and $C$ are non-coplanar. The point $M$ is the mid-point of $AC$ and $p$ denotes the plane $OAB$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">The point $R$ is such that&nbsp;$MR$ is perpendicular to $p$. Show that&nbsp;$R$ lies on a line $\ell$ with equation $\mathbf{r} = k(\mathbf{a}+\mathbf{c}) + \lambda(\mathbf{a} \times \mathbf{b})$, $\lambda \in \mathbb{R}$, where&nbsp;$k$ is a constant to be determined.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that&nbsp;$\mathbf{a}$ and $\mathbf{b}$ are unit vectors perpendicular to each other, $\mathbf{a} \cdot \mathbf{c} = -2$ and $\mathbf{b} \cdot \mathbf{c} = 4$, find in terms of $\mathbf{a}$ and $\mathbf{b}$, the position vector of the point of intersection between $\ell$ and $p$.</span><span class="inline-marks">[3]</span></div></li></ol>

For any vectors $\mathbf{m}$ and $\mathbf{n}$, explain why $\mathbf{m} \cdot (\mathbf{m} \times \mathbf{n}) = 0$.

With respect to the origin $O$, the points $A$, $B$ and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively. $O$, $A$, $B$ and $C$ are non-coplanar. The point $M$ is the mid-point of $AC$ and $p$ denotes the plane $OAB$.

The point $R$ is such that $MR$ is perpendicular to $p$. Show that $R$ lies on a line $\ell$ with equation $\mathbf{r} = k(\mathbf{a}+\mathbf{c}) + \lambda(\mathbf{a} \times \mathbf{b})$, $\lambda \in \mathbb{R}$, where $k$ is a constant to be determined.

Given that $\mathbf{a}$ and $\mathbf{b}$ are unit vectors perpendicular to each other, $\mathbf{a} \cdot \mathbf{c} = -2$ and $\mathbf{b} \cdot \mathbf{c} = 4$, find in terms of $\mathbf{a}$ and $\mathbf{b}$, the position vector of the point of intersection between $\ell$ and $p$.

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