A Math

A Math (IGCSE)

E Math

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

<div class="graph-block" data-graph-config="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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

<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

<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>
<div class="graph-block" data-graph-config="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" data-width="300" data-height="200" style="width:300px;height:200px;max-width:100%;display:block;margin:0 auto;"></div>
<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>
<div class="graph-block" data-graph-config="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" data-width="420" data-height="200" style="width:420px;height:200px;max-width:100%;display:block;margin:0 auto;"></div>
<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

Perpendicular Vectors Using the Scalar Product

2025 CJC Promo Q6

<p>With reference to the point $O$ as the origin and the $x$-$y$ plane as a horizontal plane, the pyramid $OPQRV$ has a parallelogram base $OPQR$ and height $OV$. The position vectors of the points $P$ and $R$ are $-3\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$ and $5\mathbf{i} - 2\mathbf{j} - \mathbf{k}$ respectively.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Find the coordinates of the point $S$ that lies on the line $PR$ such that the distance from $O$ to $S$ is a minimum.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Find the cartesian equations of the planes such that the perpendicular distance from each plane to the base $OPQR$ is $2\sqrt{86}$ units.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Find the acute angle between $OV$ and the vertical.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">Given that&nbsp;$\overrightarrow{QV}$ is parallel to the vector $-5\mathbf{j} + 8\mathbf{k}$, find the position vector of the point $V$. Hence find the exact volume of the pyramid $OPQRV$.<span class="inline-marks">[5]</span></div></li></ol>
<p>$\text{[Volume of a pyramid} = \frac{1}{3} \times \text{base area} \times \text{height]}$</p>

With reference to the point $O$ as the origin and the $x$-$y$ plane as a horizontal plane, the pyramid $OPQRV$ has a parallelogram base $OPQR$ and height $OV$. The position vectors of the points $P$ and $R$ are $-3\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}$ and $5\mathbf{i} - 2\mathbf{j} - \mathbf{k}$ respectively.

$\text{[Volume of a pyramid} = \frac{1}{3} \times \text{base area} \times \text{height]}$

Perpendicular Projection and Reflection in a Line

Angles Between Two Lines

Intersection of Two Lines

Three-Dimensional Geometry

Areas of Triangles and Parallelograms Using Vectors

2025 VJC P2 Q3

<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>The position vectors of the points $A$ and $B$ are $\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ and $2\mathbf{i}+3\mathbf{j}+p\mathbf{k}$ respectively, where&nbsp;$p$ is a constant. The point $C$ is such that&nbsp;$OABC$ is a rectangle, where&nbsp;$O$ is the origin.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Show that&nbsp;$p=2$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Write down the position vector of $C$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find a vector equation of the line $BC$.</span></div></li></ol>
<p>The equation of line $l$ is given by $\frac{x-1}{3}=\frac{y-1}{2}$, $z=1$.</p>
<ol class="i-list" start="4"><li><div class="list-item-content"><span class="list-item-text">Show that the lines $BC$ and $l$ are skew.</span></div></li></ol>

The position vectors of the points $A$ and $B$ are $\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ and $2\mathbf{i}+3\mathbf{j}+p\mathbf{k}$ respectively, where $p$ is a constant. The point $C$ is such that $OABC$ is a rectangle, where $O$ is the origin.

Find a vector equation of the line $BC$.

The equation of line $l$ is given by $\frac{x-1}{3}=\frac{y-1}{2}$, $z=1$.

Show that the lines $BC$ and $l$ are skew.

Equations of Lines

Relationships Between Two Lines

2006 JJC P1 Q12

<p>Relative to the origin $O$, the points $A$ and $C$ have position vectors $\mathbf{a}$ and $2\mathbf{b}$ such that</p>
<p style="text-align: center">\[\mathbf{a}=5p\mathbf{i}-2p\mathbf{j}+4p\mathbf{k} \text{ and}\hspace{0.5em} \mathbf{b}=\mathbf{i}-2\mathbf{j}+2\mathbf{k},\]</p>
<p>where&nbsp;$p$ is a positive constant. It is given that&nbsp;$|\mathbf{a}|=2|\mathbf{b}|$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Find the exact value of $p$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Evaluate $(\mathbf{a}+2\mathbf{b}) \cdot (\mathbf{a}-2\mathbf{b})$.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Evaluate $\mathbf{a} \times \mathbf{b}$ and hence find the area of triangle $OAC$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Use a geometrical reason to explain why $|\mathbf{a} \times \mathbf{b}| = \frac{1}{4}(|\mathbf{a}+2\mathbf{b}|)(|\mathbf{a}-2\mathbf{b}|)$.</span><span class="inline-marks">[2]</span></div></li></ol>

Relative to the origin $O$, the points $A$ and $C$ have position vectors $\mathbf{a}$ and $2\mathbf{b}$ such that

\[\mathbf{a}=5p\mathbf{i}-2p\mathbf{j}+4p\mathbf{k} \text{ and } \mathbf{b}=\mathbf{i}-2\mathbf{j}+2\mathbf{k},\]

where $p$ is a positive constant. It is given that $|\mathbf{a}|=2|\mathbf{b}|$.

Evaluate $(\mathbf{a}+2\mathbf{b}) \cdot (\mathbf{a}-2\mathbf{b})$.

Evaluate $\mathbf{a} \times \mathbf{b}$ and hence find the area of triangle $OAC$.

Use a geometrical reason to explain why $|\mathbf{a} \times \mathbf{b}| = \frac{1}{4}(|\mathbf{a}+2\mathbf{b}|)(|\mathbf{a}-2\mathbf{b}|)$.

Vector (Cross) Product

Scalar (Dot) Product

2025 RI P1 Q3

<p>In the diagram, $OACB$ is a trapezium where&nbsp;$AC$ is parallel to $OB$. The line $OA$ is produced to the point $D$ such that&nbsp;$\frac{OA}{AD} = \frac{1}{2}$.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Given that&nbsp;$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$, express, as simply as possible, in terms of $\mathbf{a}$ and/or $\mathbf{b}$,</div><ol class="i-list"><li><div class="list-item-paragraph">$\overrightarrow{BD}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{OC}$.</div></li></ol></li><li><div class="list-item-paragraph">Given that&nbsp;$\overrightarrow{OE} = 3\mathbf{a} + 2\mathbf{b}$,</div><ol class="i-list"><li><div class="list-item-paragraph">state the name of the quadrilateral $ODEB$,</div></li><li><div class="list-item-paragraph">explain why $O$, $C$ and $E$ lie in a straight line.</div></li></ol></li><li><div class="list-item-paragraph">Find, giving your answers as fractions in the simplest form,</div><ol class="i-list"><li><div class="list-item-paragraph">$\frac{\text{area of}\hspace{0.5em} \Delta ADC}{\text{area of}\hspace{0.5em} \Delta ODB}$,</div></li><li><div class="list-item-paragraph">$\frac{\text{area of}\hspace{0.5em} \Delta ADC}{\text{area of quadrilateral}\hspace{0.5em} ODEB}$.</div></li></ol></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><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/1768038939753-seoov8toh7.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMzg5Mzk3NTMtc2Vvb3Y4dG9oNy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0NzkwLCJleHAiOjE3OTM1MDUwOTB9._0YJK469dAX4-1XiR2WRsNv33PdsSmuWc6M4zvWxJ2E" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, $OACB$ is a trapezium where $AC$ is parallel to $OB$. The line $OA$ is produced to the point $D$ such that $\frac{OA}{AD} = \frac{1}{2}$.

Vector Subtraction

Area Ratios Using Vectors

Vector Geometry of Parallelograms and Trapezia

<p>In the diagram, $OABC$ is a parallelogram. $D$ is the mid-point of $BC$. It is given that&nbsp;$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $E$ is the point on $OC$ such that&nbsp;$5OE = 2OC$. $F$ is the point on $OD$ such that&nbsp;$\overrightarrow{OF} = \frac{1}{2}\overrightarrow{FD}$.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Express, as simply as possible, in terms of $\mathbf{a}$ and/or $\mathbf{c}$,</div><ol class="i-list"><li><div class="list-item-paragraph">$\overrightarrow{EC}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{OD}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{OF}$.</div></li></ol></li><li><div class="list-item-paragraph">Show that&nbsp;$\overrightarrow{AE} = 6\overrightarrow{FE}$.</div></li><li><div class="list-item-paragraph">Given that the coordinates of $F$ are $(4,2)$ and $\overrightarrow{AE} = \begin{pmatrix} 3 \\ -12 \end{pmatrix}$.</div><ol class="i-list"><li><div class="list-item-paragraph">Calculate $|\overrightarrow{AE}|$.</div></li><li><div class="list-item-paragraph">Find the coordinates of $E$.</div></li><li><div class="list-item-paragraph">Find the value of $t$, given that the vector $\begin{pmatrix} 4.5 \\ t \end{pmatrix}$ is parallel to $\overrightarrow{AE}$.</div></li></ol></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><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/1768037372726-zd47zjcpps.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMzczNzI3MjYtemQ0N3pqY3Bwcy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0NzkwLCJleHAiOjE3OTM1MDUwOTB9.bBdDXB2QmNzm_IAwAcgmfdipyVjIxQgYIKjZc06NqcU" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram, $OABC$ is a parallelogram. $D$ is the mid-point of $BC$. It is given that $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. $E$ is the point on $OC$ such that $5OE = 2OC$. $F$ is the point on $OD$ such that $\overrightarrow{OF} = \frac{1}{2}\overrightarrow{FD}$.

Parallel Vectors

Position Vectors

Vector Properties in Geometric Shapes: The Parallelogram

<p>The $1$ unit by $1$ unit square grid below shows the positions of point $A$, $B$ and $P$.</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/1768014954667-x9hxq2rvr68.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMTQ5NTQ2NjcteDloeHEycnZyNjgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDc5MCwiZXhwIjoxNzkzNTA1MDkwfQ.lHN_e-B7ZQkXYR94nANeB7q1pRQIqu_3i7iIPNQulIM" style="max-width: min(450px, 100%); height: auto; display: block;" alt="A coordinate grid showing the relative positions of points P, B, and A." class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Express $\overrightarrow{AB}$ and $\overrightarrow{PB}$ as column vectors.</span></div></li><li><div class="list-item-content"><span class="list-item-text">$Q$ is a point such that&nbsp;$ABQP$ is a parallelogram. Express $\overrightarrow{BQ}$ as a column vector.</span></div></li><li><div class="list-item-content"><span class="list-item-text">$R$ is a point such that&nbsp;$ABPR$ is a parallelogram. Express $\overrightarrow{BR}$ as a column vector.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Do the two vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ have the same magnitude? Is $\overrightarrow{PQ} = \overrightarrow{PR}$? Explain.</span></div></li></ol>

The $1$ unit by $1$ unit square grid below shows the positions of point $A$, $B$ and $P$.

Express $\overrightarrow{AB}$ and $\overrightarrow{PB}$ as column vectors.

$Q$ is a point such that $ABQP$ is a parallelogram. Express $\overrightarrow{BQ}$ as a column vector.

$R$ is a point such that $ABPR$ is a parallelogram. Express $\overrightarrow{BR}$ as a column vector.

Do the two vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ have the same magnitude? Is $\overrightarrow{PQ} = \overrightarrow{PR}$? Explain.

Equal Vectors

Vector Properties of Parallelograms on a Square Grid

<p>The $1$ unit by $1$ unit square grid below shows the positions of point $A$, $B$ and $P$.</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/1768014503595-lr3o66lkpb.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMTQ1MDM1OTUtbHIzbzY2bGtwYi5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0NzkwLCJleHAiOjE3OTM1MDUwOTB9.LssqvHZ3gOzL8CPxjpvF5DghHYdFnQJp7rrEiA7lGoc" style="max-width: min(450px, 100%); height: auto; display: block;" alt="A coordinate grid showing points P, A, and B." class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Express $\overrightarrow{AB}$ and $\overrightarrow{PB}$ as column vectors.</span></div></li><li><div class="list-item-content"><span class="list-item-text">$Q$ is a point such that&nbsp;$ABQP$ is a parallelogram. Express $\overrightarrow{BQ}$ as a column vector.</span></div></li><li><div class="list-item-content"><span class="list-item-text">$R$ is a point such that&nbsp;$ABPR$ is a parallelogram. Express $\overrightarrow{BR}$ as a column vector.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Do the two vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ have the same magnitude? Is $\overrightarrow{PQ} = \overrightarrow{PR}$? Explain.</span></div></li></ol>

Vector Geometry within a Cartesian Plane

<p>Relative to the origin $O$, two points $A$ and $B$ have position vectors given by $\mathbf{a}=2\mathbf{i}-\mathbf{j}$ and $\mathbf{b}=\alpha \mathbf{k}$ respectively, where&nbsp;$\alpha $ is a negative real constant.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Find the angle between $OA$ and $OB$.<span class="inline-marks">[1]</span></div></li></ol>
<p>Let&nbsp;$\angle OBA=\theta $, where&nbsp;$\theta $ is in degrees. It is given that&nbsp;$\sin \theta =\frac{1}{2}$.</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">Find the two possible values of $\theta $, and justify which of the two values is the correct value of $\theta $.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">By considering a suitable scalar product, find the exact value of $\alpha $.<span class="inline-marks">[3]</span></div></li></ol>
<p>A point $C$ has position vector $\mathbf{c}$ such that&nbsp;$\mathbf{c}=\mathbf{b}-\mathbf{a}$.</p>
<ol class="abc-list" start="4"><li><div class="list-item-paragraph">State the shape of the quadrilateral $OABC$, justifying your answer. Find the exact area of the quadrilateral $OABC$.<span class="inline-marks">[3]</span></div></li></ol>

Relative to the origin $O$, two points $A$ and $B$ have position vectors given by $\mathbf{a}=2\mathbf{i}-\mathbf{j}$ and $\mathbf{b}=\alpha \mathbf{k}$ respectively, where $\alpha $ is a negative real constant.

Let $\angle OBA=\theta $, where $\theta $ is in degrees. It is given that $\sin \theta =\frac{1}{2}$.

A point $C$ has position vector $\mathbf{c}$ such that $\mathbf{c}=\mathbf{b}-\mathbf{a}$.

2023 HCI P1 Q8

<p>$OABC$ is a rhombus. The points $A$ and $C$ are such that $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OC}=\mathbf{c}$. The point $M$ is the centre of the rhombus and angle $AOC$ is ${{60}^{\circ }}$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">By using scalar products, show that the length of projection of $\overrightarrow{OM}$ on $\overrightarrow{OC}$ is $\frac{3}{4}\left| \mathbf{c} \right|$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">By using a vector product, show that the area of triangle $OMC$ can be written as $k{{\left| c \right|}^{2}}$ where $k$ is a constant to be found.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">The point $D$ lies on $OC$ produced such that $OC:CD=2:3$. Show that the shortest distance from $D$ to the line $OA$ can be written as $t\left| \mathbf{c} \right|$ where $t$ is a constant to be found.</div><p style="text-align: right;">[3]</p></li></ol>

$OABC$ is a rhombus. The points $A$ and $C$ are such that $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OC}=\mathbf{c}$. The point $M$ is the centre of the rhombus and angle $AOC$ is ${{60}^{\circ }}$.

2012 ACJC P1 Q11

<p>Referred to an origin, $O$, the position vectors of two points $A$ and $B$ are $\mathbf{a}$ and $\mathbf{b}$ respectively. $\mathbf{a}$ and $\mathbf{b}$ are not parallel.</p>
<ol class="i-list"><li><div class="list-item-paragraph">The point $C$ lies on $AB$ produced such that $AB:BC$ is $1:3$. Find the position vector of $C$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">The point $D$ lies on $OB$ produced such that $OB:OD$ is $1:k$. Given also that $CD$ is perpendicular to $AB$, show that $k=\frac{4{{\left| \mathbf{b} \right|}^{2}}-7\mathbf{a}\cdot \mathbf{b}+3{{\left| \mathbf{a} \right|}^{2}}}{{{\left| \mathbf{b} \right|}^{2}}-\mathbf{a}\cdot \mathbf{b}}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Show that $OADC$ cannot be a parallelogram.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the area of triangle $ABD$ in terms of $k$.</div><p style="text-align: right;">[3]</p></li></ol>

Referred to an origin, $O$, the position vectors of two points $A$ and $B$ are $\mathbf{a}$ and $\mathbf{b}$ respectively. $\mathbf{a}$ and $\mathbf{b}$ are not parallel.

2013 RVHS P1 Q9

<p>The position vectors of points $A$, $B$ and $C$ relative to $O$ are $\mathbf{a}$,$\mathbf{b}$ and $\mathbf{c}$ respectively, where $\mathbf{a}$,$\mathbf{b}$ and $\mathbf{c}$ are non-parallel vectors. The point $M$ is the midpoint of $AB$ and the point $N$ lies on $AC$ produced such that $AN:CN=4:3$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the position vector of $M$ and $N$, giving your answers in terms of $\mathbf{a}$,$\mathbf{b}$ and/or $\mathbf{c}$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">The point $D$ is such that $ABDN$ is a parallelogram. Find the position vector of $D$ in terms of vectors $\mathbf{a}$,$\mathbf{b}$ and $\mathbf{c}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">The line $MN$ and $AD$ intersect at point $Q$. Find the position vector of $Q$ in terms of vector of $\mathbf{a}$,$\mathbf{b}$ and $\mathbf{c}$.</div><p style="text-align: right;">[5]</p></li></ol>

The position vectors of points $A$, $B$ and $C$ relative to $O$ are $\mathbf{a}$,$\mathbf{b}$ and $\mathbf{c}$ respectively, where $\mathbf{a}$,$\mathbf{b}$ and $\mathbf{c}$ are non-parallel vectors. The point $M$ is the midpoint of $AB$ and the point $N$ lies on $AC$ produced such that $AN:CN=4:3$.

2022 TJC J1 MYE Q8

<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/1779196427808-ga54r0gz7k6.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzkxOTY0Mjc4MDgtZ2E1NHIwZ3o3azYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDc5MCwiZXhwIjoxNzkzNTA1MDkwfQ.bBJbUE4XO-FGc4FbxmBOYZn0AvsDj4Zvfouj3EVTIS0" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>$OACB$ is a square with origin $O$. Points $A$ and $B$ are such that&nbsp;$\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. $M$ and $N$ are midpoints of $OB$ and $AC$ respectively. $S$ is on $AC$ such that&nbsp;$AS:AC=1:4$. $MS$ produced and $BN$ produced meet at point $T$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Find the position vector $\overrightarrow{OS}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-paragraph">Find vector equations of the line $MS$ and the line $BN$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">By finding the position vector of $T$, deduce that&nbsp;$T$, $A$ and $O$ are collinear points.<span class="inline-marks">[2]</span></div></li></ol>

$OACB$ is a square with origin $O$. Points $A$ and $B$ are such that $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. $M$ and $N$ are midpoints of $OB$ and $AC$ respectively. $S$ is on $AC$ such that $AS:AC=1:4$. $MS$ produced and $BN$ produced meet at point $T$.

2023 RVHS P1 Q5

<p>Relative to the origin $O$, the position vectors of the fixed points $A$ and $B$ are $\mathbf{a}$ and $\mathbf{b}$ respectively.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">It is given that&nbsp;$|\mathbf{a}|=1$, $|\mathbf{b}|=2$ and $|\mathbf{3b}-\mathbf{2a}|=\sqrt{34}$.</div><ol class="i-list"><li><div class="list-item-paragraph">By considering the scalar product $(3\mathbf{b}-2\mathbf{a})\cdot (3\mathbf{b}-2\mathbf{a})$, find the value of $\mathbf{a}\cdot \mathbf{b}$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Hence find the exact value of $|\mathbf{a}\times \mathbf{b}|$ and give a geometrical meaning of this value.<span class="inline-marks">[3]</span></div></li></ol></li><li><div class="list-item-paragraph">A point $P$ divides $AB$ in the ratio $2:1$ and a point $M$ is the mid-point of $BC$, where the point $C$ has position vector $\mathbf{c}$. It is given that the point $N$ is such that&nbsp;$OBMN$ forms a rhombus and $P$ is also the mid-point of $MN$. Show that&nbsp;$2\mathbf{a}+4\mathbf{b}-3\mathbf{c}=\mathbf{0}$.<span class="inline-marks">[3]</span></div></li></ol>

Relative to the origin $O$, the position vectors of the fixed points $A$ and $B$ are $\mathbf{a}$ and $\mathbf{b}$ respectively.

2023 MI P1 Q8

<p>The points $P$ and $Q$ have position vectors $\mathbf{a}+\mathbf{b}$ and $3\mathbf{a}-2\mathbf{b}$ respectively relative to the origin $O$.</p>
<p>Given that $OPQR$ is a parallelogram, express the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.</p>
<p>By evaluating two scalar products, show that if $OPQR$ is a square, then ${{\left| \mathbf{a} \right|}^{2}}=2{{\left| \mathbf{b} \right|}^{2}}$.</p>

The points $P$ and $Q$ have position vectors $\mathbf{a}+\mathbf{b}$ and $3\mathbf{a}-2\mathbf{b}$ respectively relative to the origin $O$.

Given that $OPQR$ is a parallelogram, express the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

By evaluating two scalar products, show that if $OPQR$ is a square, then ${{\left| \mathbf{a} \right|}^{2}}=2{{\left| \mathbf{b} \right|}^{2}}$.

Parallelogram and Square

<p>$ABCD$ is a square, and $M$ and $N$ are the midpoints of $BC$ and $CD$ respectively. Express $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{BD}$ in terms of $\mathbf{p}$ and $\mathbf{q}$, where $\overrightarrow{AM}=\mathbf{p}$ and $\overrightarrow{AN}=\mathbf{q}$.</p>

$ABCD$ is a square, and $M$ and $N$ are the midpoints of $BC$ and $CD$ respectively. Express $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{BD}$ in terms of $\mathbf{p}$ and $\mathbf{q}$, where $\overrightarrow{AM}=\mathbf{p}$ and $\overrightarrow{AN}=\mathbf{q}$.

Vector Algebra

EJC Vectors I Tutorial Q8

<p>Relative to the origin $O$, points $A$ and $B$ have position vectors $4\mathbf{i}+3\mathbf{j}$ and $5\mathbf{i}+7\mathbf{j}$ respectively. The line ${{l}_{1}}$ passes through point $B$ and is parallel to the vector $\mathbf{i}+2\mathbf{j}+\alpha \mathbf{k}$. The line ${{l}_{2}}$ has cartesian equation $\frac{1-x}{2}=\frac{z-2}{1}$, $y=4$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Given that ${{l}_{1}}$ and ${{l}_{2}}$ are perpendicular, show that the value of $\alpha $ is $2$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">Find the length of projection of $\overrightarrow{BA}$ onto line ${{l}_{1}}$. Hence find the vector $\overrightarrow{BN}$, where $N$ is the foot of perpendicular from $A$ to ${{l}_{1}}$.</div><p style="text-align: right;">[4]</p></li><li><div class="list-item-paragraph">Point $A'$ is the reflection of point $A$ in the line ${{l}_{1}}$ and point $C$ is such that $AA'CB$ forms a parallelogram. Find the exact area of the parallelogram $AA'CB$.</div><p style="text-align: right;">[3]</p></li></ol>

Relative to the origin $O$, points $A$ and $B$ have position vectors $4\mathbf{i}+3\mathbf{j}$ and $5\mathbf{i}+7\mathbf{j}$ respectively. The line ${{l}_{1}}$ passes through point $B$ and is parallel to the vector $\mathbf{i}+2\mathbf{j}+\alpha \mathbf{k}$. The line ${{l}_{2}}$ has cartesian equation $\frac{1-x}{2}=\frac{z-2}{1}$, $y=4$.

Math Question Bank