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="300" data-height="200" data-proportional style="width:300px;height:200px;max-width:100%;display:flow-root;margin:0 auto;margin-bottom:0px;--graph-width:300px;--graph-aspect:300 / 200;"></div>
<p>With reference to the origin $O$, the points $A$, $B$ and $C$ are such that vectors $\overrightarrow{OA}=\mathbf{a}$, $\overrightarrow{OB}=\mathbf{b}$ and $\overrightarrow{OC}=2\mathbf{a}-2\mathbf{b}$.</p>
<ol class="abc-list"><li data-list-item-id="acjc-q08-item01" data-list-item-path="acjc-q08-item01"><div class="list-item-paragraph">Given that&nbsp;$|\mathbf{a}|=2$, $|\mathbf{b}|=3$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ is $\frac{\pi}{6}$, find the area of triangle $OBC$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="acjc-q08-item02" data-list-item-path="acjc-q08-item02"><div class="list-item-paragraph">The lines $OA$ and $BC$ intersect at point $X$.</div><div class="list-item-paragraph">Show that&nbsp;$\displaystyle \overrightarrow{OX}=\frac23\mathbf{a}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="acjc-q08-item03" data-list-item-path="acjc-q08-item03"><div class="list-item-paragraph">The plane $p$ has equation $\mathbf{r}\cdot\mathbf{n}=d$ and contains the point $A$, where&nbsp;$\mathbf{n}$ is a unit vector.</div><div class="list-item-paragraph">Find the shortest distance from point $X$ to plane $p$, in the form of $\frac1k|d|$, where&nbsp;$k$ is a positive integer to be determined.<span class="inline-marks">[3]</span></div></li></ol>

With reference to the origin $O$, the points $A$, $B$ and $C$ are such that vectors $\overrightarrow{OA}=\mathbf{a}$, $\overrightarrow{OB}=\mathbf{b}$ and $\overrightarrow{OC}=2\mathbf{a}-2\mathbf{b}$.

Given that $|\mathbf{a}|=2$, $|\mathbf{b}|=3$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ is $\frac{\pi}{6}$, find the area of triangle $OBC$.

Distance from a Point to a Plane

Intersection of Two Lines

Areas of Triangles and Parallelograms Using Vectors

2026 ACJC P1 Q8

Vectors

<p>Relative to the origin $O$, two points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. It is given that&nbsp;$|\mathbf{a}|=2$, $|\mathbf{b}|=1$ and $|3\mathbf{a}-2\mathbf{b}|=\sqrt{37}$.</p>
<ol class="i-list"><li data-list-item-id="h2-2025-sajc-promo-q8-sajc-q8-item-block-1-item-0" data-list-item-path="h2-2025-sajc-promo-q8-sajc-q8-item-block-1-item-0"><div class="list-item-paragraph">By considering the scalar product $(3\mathbf{a}-2\mathbf{b})\cdot(3\mathbf{a}-2\mathbf{b})$, show that&nbsp;$\mathbf{a}\cdot\mathbf{b}=\dfrac{1}{4}$, and give the geometrical meaning of $|\mathbf{a}\cdot\mathbf{b}|$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="h2-2025-sajc-promo-q8-sajc-q8-item-block-1-item-1" data-list-item-path="h2-2025-sajc-promo-q8-sajc-q8-item-block-1-item-1"><div class="list-item-paragraph">By finding the cosine of the angle between vectors $\mathbf{a}$ and $\mathbf{b}$, deduce the exact value of $|(\mathbf{a}-\mathbf{b})\times\mathbf{b}|$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-sajc-promo-q8-sajc-q8-item-block-1-item-2" data-list-item-path="h2-2025-sajc-promo-q8-sajc-q8-item-block-1-item-2"><div class="list-item-paragraph">Interpret geometrically the value of $|(\mathbf{a}-\mathbf{b})\times\mathbf{b}|$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-2025-sajc-promo-q8-sajc-q8-item-block-1-item-3" data-list-item-path="h2-2025-sajc-promo-q8-sajc-q8-item-block-1-item-3"><div class="list-item-paragraph">Write down, in terms of $\mathbf{a}$ and $\mathbf{b}$, a vector equation of the line that passes through $O$ and bisects the angle $AOB$.<span class="inline-marks">[2]</span></div></li></ol>

Relative to the origin $O$, two points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. It is given that $|\mathbf{a}|=2$, $|\mathbf{b}|=1$ and $|3\mathbf{a}-2\mathbf{b}|=\sqrt{37}$.

By considering the scalar product $(3\mathbf{a}-2\mathbf{b})\cdot(3\mathbf{a}-2\mathbf{b})$, show that $\mathbf{a}\cdot\mathbf{b}=\dfrac{1}{4}$, and give the geometrical meaning of $|\mathbf{a}\cdot\mathbf{b}|$.

By finding the cosine of the angle between vectors $\mathbf{a}$ and $\mathbf{b}$, deduce the exact value of $|(\mathbf{a}-\mathbf{b})\times\mathbf{b}|$.

Interpret geometrically the value of $|(\mathbf{a}-\mathbf{b})\times\mathbf{b}|$.

Write down, in terms of $\mathbf{a}$ and $\mathbf{b}$, a vector equation of the line that passes through $O$ and bisects the angle $AOB$.

Projection Lengths

Angles Between Vectors

Angle Bisectors Using Vectors

Equations of Lines

Vector (Cross) Product

Scalar (Dot) Product

2025 SAJC Promo Q8

<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

Three-Dimensional Geometry

Equations of Planes

Vectors in a Parallelogram

2025 VJC P2 Q3

<p>The position vectors of points $A$, $B$ and $C$, relative to the origin $O$ are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively, where</p>
<p style="text-align: center">\[\mathbf{a} = 3\mathbf{i} + 2\mathbf{j} - \mathbf{k},\quad \mathbf{b} = 4\mathbf{i} - 3\mathbf{j} + 2\mathbf{k},\quad \mathbf{c} = 3\mathbf{i} - \mathbf{j} - \mathbf{k}.\]</p>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-paragraph">Find $\mathbf{a} \times \mathbf{b}$ and deduce the exact area of triangle $OAB$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b"><div class="list-item-paragraph">Hence, find the volume of the tetrahedron $OABC$.</div><div class="list-item-paragraph">[The volume of a tetrahedron is $\frac{1}{3} \times \text{base area} \times \text{perpendicular height}$.]<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c"><div class="list-item-paragraph">Find the perpendicular distance from $B$ to line $OA$.<span class="inline-marks">[2]</span></div></li></ol>

The position vectors of points $A$, $B$ and $C$, relative to the origin $O$ are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively, where

\[\mathbf{a} = 3\mathbf{i} + 2\mathbf{j} - \mathbf{k},\quad \mathbf{b} = 4\mathbf{i} - 3\mathbf{j} + 2\mathbf{k},\quad \mathbf{c} = 3\mathbf{i} - \mathbf{j} - \mathbf{k}.\]

Distance from a Point to a Line

2024 NYJC Promo Q5

<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 $PQ : QR = 4 : 5$.

Collinearity Using Vectors

Ratio Theorem and Section Formula

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.

2024 MI PRU1 Q4

<p>In triangle $PQR$, the vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ represent the vectors $\overrightarrow{PQ}$, $\overrightarrow{QR}$ and $\overrightarrow{PR}$ respectively.</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/1779426366202-yu4qd1rzjpn.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Nzk0MjYzNjYyMDIteXU0cWQxcnpqcG4ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDc0NiwiZXhwIjoxNzkzNTA1MDQ2fQ.bgvQ2aQsBUavWT9gs4_3cq5IclVTgFw4ETejb-jBi8c" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-paragraph">State a geometrical interpretation of $|\mathbf{c} \times \mathbf{\hat{a}}|$ and hence show that the area of the triangle $PQR$ can be written as&nbsp;$\frac{1}{2}|\mathbf{c} \times \mathbf{{a}}|$ units$^2$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b"><div class="list-item-paragraph">With clear explanations, show that&nbsp;$|\mathbf{c} \times \mathbf{a}| = |\mathbf{b} \times \mathbf{a}|$ and deduce that&nbsp;$\frac{|\mathbf{c}|}{\sin \angle PQR} = \frac{|\mathbf{b}|}{\sin \angle QPR}$ (Sine Rule).<span class="inline-marks">[2]</span></div></li></ol>

In triangle $PQR$, the vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ represent the vectors $\overrightarrow{PQ}$, $\overrightarrow{QR}$ and $\overrightarrow{PR}$ respectively.

2024 ASRJC Promo Q5

<p>If&nbsp;$|\mathbf{a} \times \mathbf{b}| = 16$ and $|\mathbf{b} - \mathbf{a}| = 10$, find the shortest distance from $O$ to $AB$.<span class="inline-marks">[2]</span></p>

If $|\mathbf{a} \times \mathbf{b}| = 16$ and $|\mathbf{b} - \mathbf{a}| = 10$, find the shortest distance from $O$ to $AB$.

<p>Referred to the origin $O$, the points $A$ and $B$ have position vectors $\mathbf{a}$ and $\mathbf{b}$ such that</p>
<p style="text-align: center">\[\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k} \quad \text{and} \quad \mathbf{b} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} .\]</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find the size of angle $OAB$.</span><span class="inline-marks">[2]</span></div></li></ol>
<p>The point $C$ has position vector $\mathbf{c}$ given by $\mathbf{c} = \lambda \mathbf{a} + \mu \mathbf{b}$, where&nbsp;$\lambda$ and $\mu$ are positive constants.<br>Given that the area of triangle $OAC$ is twice that of triangle $OBC$,</p>
<ol class="i-list" start="2"><li><div class="list-item-content"><span class="list-item-text">find $\mu$ in terms of $\lambda$,</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">hence, if&nbsp;$OC = \sqrt{118}$, find the position vector $\mathbf{c}$.</span><span class="inline-marks">[4]</span></div></li></ol>

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

\[\mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k} \quad \text{and} \quad \mathbf{b} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} .\]

The point $C$ has position vector $\mathbf{c}$ given by $\mathbf{c} = \lambda \mathbf{a} + \mu \mathbf{b}$, where $\lambda$ and $\mu$ are positive constants.
Given that the area of triangle $OAC$ is twice that of triangle $OBC$,

hence, if $OC = \sqrt{118}$, find the position vector $\mathbf{c}$.

Magnitude of a Vector

2021 JPJC P1 Q2

<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>Relative to the origin $O$, the position vectors of $A$, $B$ and $C$  are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively where&nbsp;$\mathbf{c}=3\mathbf{a}-5\mathbf{b}$. Given that&nbsp;$\left| \mathbf{a} \right|=\sqrt{6}$, $\mathbf{b}$ is a unit vector and the angle between $\mathbf{a}$ and $\mathbf{b}$ is $\frac{2}{3}\pi $, find the exact area of triangle $ABC$.</p>

Relative to the origin $O$, the position vectors of $A$, $B$ and $C$  are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively where $\mathbf{c}=3\mathbf{a}-5\mathbf{b}$. Given that  $\left| \mathbf{a} \right|=\sqrt{6}$, $\mathbf{b}$ is a unit vector and the angle between $\mathbf{a}$ and $\mathbf{b}$ is $\frac{2}{3}\pi $, find the exact area of triangle $ABC$.

<p>Relative to the origin $O$, the position vectors of $A$, $B$ and $C$  are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively where&nbsp;$\mathbf{c}=5\mathbf{a}-\mathbf{b}$. Given that&nbsp;$\mathbf{a}$ is a unit vector, $\left| \mathbf{b} \right|=\sqrt{3}$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ is $\frac{1}{3}\pi $, find the exact area of triangle $ABC$.</p>

Relative to the origin $O$, the position vectors of $A$, $B$ and $C$  are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively where $\mathbf{c}=5\mathbf{a}-\mathbf{b}$. Given that $\mathbf{a}$ is a unit vector, $\left| \mathbf{b} \right|=\sqrt{3}$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ is $\frac{1}{3}\pi $, find the exact area of triangle $ABC$.

<p>The planes $p_1$, $p_2$ and $p_3$ have equations \[p_1 : -x + 2y + 4z = 5,\]\[p_2 : 3x + ay - 6z = 6,\]\[p_3 : 2x - 4y - 8z = b,\]where&nbsp;$a$ and $b$ are real constants, $b &gt; 0$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">The line $l$ has equation $\frac{x - 5}{8} = \frac{3 - y}{6} = \frac{z - 1}{5}$. Given that the planes $p_1$ and $p_2$ intersect at $l$, show that&nbsp;$a = -1$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the acute angle between $p_1$ and $p_2$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">The plane $p_4$ has equation $\mathbf{r} \cdot \begin{pmatrix} 1 \\ -2 \\ -4 \end{pmatrix} = 0$. Given that&nbsp;$p_1$ and $p_3$ are equidistant from $p_4$, find the value of $b$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">The point $A$ with coordinates $(-1, 0, 1)$ lies on $p_1$, and $F$ is the foot of the perpendicular from $A$ to $l$. The point $B$ is the reflection of $F$ in the plane $p_3$. Find the length of $AF$ and hence find the area of triangle $AFB$.</span><span class="inline-marks">[5]</span></div></li></ol>

The planes $p_1$, $p_2$ and $p_3$ have equations \[p_1 : -x + 2y + 4z = 5,\]\[p_2 : 3x + ay - 6z = 6,\]\[p_3 : 2x - 4y - 8z = b,\]where $a$ and $b$ are real constants, $b > 0$.

The line $l$ has equation $\frac{x - 5}{8} = \frac{3 - y}{6} = \frac{z - 1}{5}$. Given that the planes $p_1$ and $p_2$ intersect at $l$, show that $a = -1$.

Find the acute angle between $p_1$ and $p_2$.

The plane $p_4$ has equation $\mathbf{r} \cdot \begin{pmatrix} 1 \\ -2 \\ -4 \end{pmatrix} = 0$. Given that $p_1$ and $p_3$ are equidistant from $p_4$, find the value of $b$.

The point $A$ with coordinates $(-1, 0, 1)$ lies on $p_1$, and $F$ is the foot of the perpendicular from $A$ to $l$. The point $B$ is the reflection of $F$ in the plane $p_3$. Find the length of $AF$ and hence find the area of triangle $AFB$.

Perpendicular Projection and Reflection in a Plane

Relationships Between a Line and a Plane

Angle Between Two Planes

2018 HCI CT Q8

<p>Planes $p$ and $q$ are perpendicular. Plane $p$ has equation $\mathbf{r} \cdot (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 6$. Plane $q$ contains the line $l$ with equation $\mathbf{r} = \mathbf{j} - 12\mathbf{k} + \lambda(4\mathbf{i} + \mathbf{j} + 10\mathbf{k})$ where&nbsp;$\lambda$ is a parameter. The point $A$ has coordinates $(0, 1, -12)$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Find a cartesian equation of $q$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find a vector equation of the line $m$, where&nbsp;$p$ and $q$ meet.</span><span class="inline-marks">[2]</span></div></li></ol>
<p>$l$ passes through $p$ at point $B$. Point $C$ is the foot of the perpendicular of $A$ to $p$.</p>
<ol class="abc-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Find the position vector of $B$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the position vector of $C$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the exact area of $\Delta ABC$.</span><span class="inline-marks">[2]</span></div></li></ol>

Planes $p$ and $q$ are perpendicular. Plane $p$ has equation $\mathbf{r} \cdot (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 6$. Plane $q$ contains the line $l$ with equation $\mathbf{r} = \mathbf{j} - 12\mathbf{k} + \lambda(4\mathbf{i} + \mathbf{j} + 10\mathbf{k})$ where $\lambda$ is a parameter. The point $A$ has coordinates $(0, 1, -12)$.

Find a vector equation of the line $m$, where $p$ and $q$ meet.

$l$ passes through $p$ at point $B$. Point $C$ is the foot of the perpendicular of $A$ to $p$.

Intersection of a Line and a Plane

Line of Intersection of Two Planes

2025 YIJC P1 Q10

<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>The points $A$, $B$ and $A'$ lie on a circle with center $O$ such that&nbsp;$\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. It is also given that the line segment $AA'$ is a straight line that passes through $O$ (see diagram).</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/1769240658720-er0ri3x0a3l.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjkyNDA2NTg3MjAtZXIwcmkzeDBhM2wucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDc0NiwiZXhwIjoxNzkzNTA1MDQ2fQ.aaGRb_Dc2_leeppCFRYDHnn1iekkdR9utUFkGkzLTbw" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Using a suitable scalar product, prove that&nbsp;$\angle ABA'=90{}^\circ$.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Express the area of $\Delta ABA'$ in the form $k|\mathbf{a} \times \mathbf{b}|$, where&nbsp;$k$ is a real constant.</span><span class="inline-marks">[2]</span></div></li></ol>

The points $A$, $B$ and $A'$ lie on a circle with center $O$ such that $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. It is also given that the line segment $AA'$ is a straight line that passes through $O$ (see diagram).

Using a suitable scalar product, prove that $\angle ABA'=90{}^\circ$.

Express the area of $\Delta ABA'$ in the form $k|\mathbf{a} \times \mathbf{b}|$, where $k$ is a real constant.

Perpendicular Vectors Using the Scalar Product

2025 SAJC P1 Q11

<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 $|\mathbf{a} \times \mathbf{b}| = \frac{1}{4}(|\mathbf{a}+2\mathbf{b}|)(|\mathbf{a}-2\mathbf{b}|)$.

2025 RI P1 Q3

<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-paragraph">A triangle $ABC$ is such that&nbsp;$AB^2 + BC^2 = AC^2$.<br>By considering $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$ and using the fact that&nbsp;$|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}$ for any vector $\mathbf{v}$, prove that&nbsp;$\angle ABC$ is a right-angle.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b"><div class="list-item-paragraph">In a triangle $ABC$, the point $D$ divides $AC$ in the ratio $\lambda : 1 - \lambda$, where&nbsp;$0 &lt; \lambda &lt; 1$. Let the position vectors of $A$, $B$, $C$ and $D$ be denoted by $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ and $\mathbf{d}$ respectively. Show that the area of triangle $ABD$ is given by\[k|\mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}|\]where&nbsp;$k$ is to be determined in terms of $\lambda$.<span class="inline-marks">[4]</span></div></li></ol>

2025 NYJC P1 Q3

<p>Referred to the origin $O$, the 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 vectors. The point $C$ lies on $OA$ such that&nbsp;$OC : CA = 1 : 2$. The point $D$ lies on $OB$ such that&nbsp;$OB : DB = 3 : 1$.</p>
<ol class="i-list"><li><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><div class="list-item-content"><span class="list-item-text">Show that the point $E$ where the lines $AD$ and $BC$ meet has position vector \[\frac{1}{7}\mathbf{a} + \frac{4}{7}\mathbf{b}\]</span><span class="inline-marks">[5]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Show that the area of triangle $OAE$ can be written as&nbsp;$k|\mathbf{a} \times \mathbf{b}|$, where&nbsp;$k$ is a constant to be found.</span><span class="inline-marks">[3]</span></div></li></ol>
<p>It is further given that&nbsp;$\mathbf{b}$ is a unit vector.</p>
<ol class="i-list" start="4"><li><div class="list-item-content"><span class="list-item-text">Give the geometrical meaning of $|\mathbf{a} \cdot \mathbf{b}|$.</span><span class="inline-marks">[1]</span></div></li><li><div class="list-item-content"><span class="list-item-text">If&nbsp;$|\mathbf{a} \cdot \mathbf{b}| = \frac{1}{2}|\mathbf{a}|$, find the perpendicular distance of point $A$ to $OB$, leaving your answer in terms of $|\mathbf{a}|$.</span><span class="inline-marks">[3]</span></div></li></ol>

Referred to the origin $O$, the 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 vectors. The point $C$ lies on $OA$ such that $OC : CA = 1 : 2$. The point $D$ lies on $OB$ such that $OB : DB = 3 : 1$.

Show that the point $E$ where the lines $AD$ and $BC$ meet has position vector \[\frac{1}{7}\mathbf{a} + \frac{4}{7}\mathbf{b}\]

Show that the area of triangle $OAE$ can be written as $k|\mathbf{a} \times \mathbf{b}|$, where $k$ is a constant to be found.

It is further given that $\mathbf{b}$ is a unit vector.

Give the geometrical meaning of $|\mathbf{a} \cdot \mathbf{b}|$.

If $|\mathbf{a} \cdot \mathbf{b}| = \frac{1}{2}|\mathbf{a}|$, find the perpendicular distance of point $A$ to $OB$, leaving your answer in terms of $|\mathbf{a}|$.

2025 MI P1 Q11

<p>With reference to the origin $O$, the points $A$, $P$ and $Q$ have position vectors $\mathbf{a}$, $\mathbf{p}$ and $\mathbf{q}$ respectively. A straight line $l$ through the point $A$ is parallel to a unit vector $\mathbf{e}$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">The point $Q$ lies on $l$. Show that&nbsp;$(\mathbf{q} - \mathbf{a}) \times \mathbf{e} = \mathbf{0}$.<span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Given that&nbsp;$P$ is not on $l$, give the geometrical meaning of $|(\mathbf{p} - \mathbf{a}) \times \mathbf{e}|$.</span><span class="inline-marks">[1]</span></div></li></ol>
<p>It is also given that&nbsp;$\mathbf{p} = 3\mathbf{q}$ and $AQ$ is $2$ units.</p>
<ol class="abc-list" start="3"><li><div class="list-item-paragraph">Using the results in <strong>(a)</strong> and <strong>(b)</strong>, or otherwise, show that the area of triangle $APQ$ can be written as&nbsp;$k|\mathbf{q} \times \mathbf{e}|$, where&nbsp;$k$ is a constant to be determined.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence, or otherwise, find the acute angle between $l$ and $PQ$, given further that the area of triangle $\Delta APQ$ is $3$ units${}^2$ and $|\mathbf{q}| = 2$.</span><span class="inline-marks">[3]</span></div></li></ol>

With reference to the origin $O$, the points $A$, $P$ and $Q$ have position vectors $\mathbf{a}$, $\mathbf{p}$ and $\mathbf{q}$ respectively. A straight line $l$ through the point $A$ is parallel to a unit vector $\mathbf{e}$.

Given that $P$ is not on $l$, give the geometrical meaning of $|(\mathbf{p} - \mathbf{a}) \times \mathbf{e}|$.

It is also given that $\mathbf{p} = 3\mathbf{q}$ and $AQ$ is $2$ units.

Hence, or otherwise, find the acute angle between $l$ and $PQ$, given further that the area of triangle $\Delta APQ$ is $3$ units${}^2$ and $|\mathbf{q}| = 2$.

Math Question Bank