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

<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

Distance from a Point to a Plane

Angle Between a Line and a Plane

Equations of Planes

Perpendicular Vectors Using the Scalar Product

Ratio Theorem and Section Formula

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Vectors

<ol class="abc-list"><li data-list-item-id="h2-2025-ri-promo-q6-ri-q6-item-block-0-item-0" data-list-item-path="h2-2025-ri-promo-q6-ri-q6-item-block-0-item-0"><div class="list-item-paragraph">The non-zero vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are such that&nbsp;$3\mathbf{a}\times\mathbf{b}=2\mathbf{b}\times\mathbf{c}$, where&nbsp;$3\mathbf{a}\ne -2\mathbf{c}$. Find a linear relationship between $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-ri-promo-q6-ri-q6-item-block-0-item-1" data-list-item-path="h2-2025-ri-promo-q6-ri-q6-item-block-0-item-1"><div class="list-item-paragraph"><p>The variable position vector $\mathbf{v}$ satisfies the equation</p></div><div class="list-item-paragraph"><p style="text-align: center;">$\mathbf{v}\times(-\mathbf{i}+2\mathbf{j}+\mathbf{k})=a\mathbf{i}+\mathbf{j}+\mathbf{k},$</p></div><div class="list-item-paragraph"><p>where $a$ is a non-zero constant.</p></div><ol class="i-list"><li data-list-item-id="h2-2025-ri-promo-q6-ri-q6-item-block-0-item-1-sub-0" data-list-item-path="h2-2025-ri-promo-q6-ri-q6-item-block-0-item-1/h2-2025-ri-promo-q6-ri-q6-item-block-0-item-1-sub-0"><div class="list-item-paragraph">Show that&nbsp;$a=3$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-2025-ri-promo-q6-ri-q6-item-block-0-item-1-sub-1" data-list-item-path="h2-2025-ri-promo-q6-ri-q6-item-block-0-item-1/h2-2025-ri-promo-q6-ri-q6-item-block-0-item-1-sub-1"><div class="list-item-paragraph">Hence, by finding the set of position vectors $\mathbf{v}$, describe geometrically the set of points represented by the position vectors $\mathbf{v}$.<span class="inline-marks">[3]</span></div></li></ol></li></ol>

The non-zero vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ are such that $3\mathbf{a}\times\mathbf{b}=2\mathbf{b}\times\mathbf{c}$, where $3\mathbf{a}\ne -2\mathbf{c}$. Find a linear relationship between $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$.

Parallel Vectors

Equations of Lines

Vector (Cross) Product

Geometrical Interpretation of Vector Equations

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<ol class="abc-list"><li data-list-item-id="h2-2025-nyjc-promo-q7-nyjc-q7-item-block-0-item-0" data-list-item-path="h2-2025-nyjc-promo-q7-nyjc-q7-item-block-0-item-0"><div class="list-item-paragraph">Given that&nbsp;$\mathbf{a}$ and $\mathbf{b}$ are non-zero vectors such that&nbsp;$\mathbf{a}=(\mathbf{a}\cdot\mathbf{b})\mathbf{b}$. State the relationship between $\mathbf{a}$ and $\mathbf{b}$. Find $|\mathbf{b}|$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-2025-nyjc-promo-q7-nyjc-q7-item-block-0-item-1" data-list-item-path="h2-2025-nyjc-promo-q7-nyjc-q7-item-block-0-item-1"><div class="list-item-paragraph">With reference to the origin $O$, the points $M$ and $N$ are such that&nbsp;$\overrightarrow{OM}=\alpha\mathbf{i}+4\mathbf{j}+\beta\mathbf{k}$ and $\overrightarrow{ON}=4\mathbf{i}+\alpha\mathbf{j}+\beta\mathbf{k}$, where&nbsp;$\alpha$ and $\beta$ are constants.</div><ol class="i-list"><li data-list-item-id="h2-2025-nyjc-promo-q7-nyjc-q7-item-block-0-item-1-sub-0" data-list-item-path="h2-2025-nyjc-promo-q7-nyjc-q7-item-block-0-item-1/h2-2025-nyjc-promo-q7-nyjc-q7-item-block-0-item-1-sub-0"><div class="list-item-paragraph">Given that&nbsp;$M$ lies in the $xy$-plane and $|\overrightarrow{OM}|=5$, find the values of $\alpha$ and $\beta$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-nyjc-promo-q7-nyjc-q7-item-block-0-item-1-sub-1" data-list-item-path="h2-2025-nyjc-promo-q7-nyjc-q7-item-block-0-item-1/h2-2025-nyjc-promo-q7-nyjc-q7-item-block-0-item-1-sub-1"><div class="list-item-paragraph"><p>Given <strong>instead</strong> that&nbsp;$\overrightarrow{OM}$ is perpendicular to $\overrightarrow{ON}$, find an expression for&nbsp;$|\overrightarrow{OM}\times\overrightarrow{ON}|$ in terms of $\alpha$, giving your answer in fully factorised form.</p><span class="inline-marks">[3]</span></div></li></ol></li></ol>

Given that $\mathbf{a}$ and $\mathbf{b}$ are non-zero vectors such that $\mathbf{a}=(\mathbf{a}\cdot\mathbf{b})\mathbf{b}$. State the relationship between $\mathbf{a}$ and $\mathbf{b}$. Find $|\mathbf{b}|$.

With reference to the origin $O$, the points $M$ and $N$ are such that $\overrightarrow{OM}=\alpha\mathbf{i}+4\mathbf{j}+\beta\mathbf{k}$ and $\overrightarrow{ON}=4\mathbf{i}+\alpha\mathbf{j}+\beta\mathbf{k}$, where $\alpha$ and $\beta$ are constants.

Finding Points on a Plane

Magnitude of a Vector

Scalar (Dot) Product

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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>
<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

Vectors in a Parallelogram

2025 CJC Promo Q6

<p>Two vectors $\mathbf{a}$ and $\mathbf{b}$ are such that&nbsp;$\vert{}\mathbf{a} \times \mathbf{b}\vert{} = 3$. It is given that&nbsp;$\mathbf{a}$ is a unit vector and $\mathbf{b} \cdot \mathbf{b} = 9$.</p>
<p>Show that&nbsp;$\mathbf{a}$ and $\mathbf{b}$ are perpendicular.<span class="inline-marks">[3]</span></p>

Two vectors $\mathbf{a}$ and $\mathbf{b}$ are such that $\vert{}\mathbf{a} \times \mathbf{b}\vert{} = 3$. It is given that $\mathbf{a}$ is a unit vector and $\mathbf{b} \cdot \mathbf{b} = 9$.

Show that $\mathbf{a}$ and $\mathbf{b}$ are perpendicular.

N2024 P1 Q2

<p>The planes $\pi_1$ and $\pi_2$, which meet in the line $l_1$, have equations</p>
<p style="text-align: center">\[\pi_1 : \mathbf{r} \cdot \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = 25 \quad \text{and} \quad \pi_2 : x + ky - 2z = -15,\]</p>
<p>where&nbsp;$k$ is a constant.</p>
<p>Another line $l_2$ has equation $\mathbf{r} = \begin{pmatrix} -15 \\ 0 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}, \beta \in \mathbb{R}$.</p>
<ol class="i-list"><li data-list-item-id="imported-item-block-4-item-0" data-list-item-path="imported-item-block-4-item-0"><div class="list-item-paragraph">Determine the position vector of the point on $l_1$ such that the coordinates of this point is independent of $k$. Hence find a vector equation of $l_1$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="imported-item-block-4-item-1" data-list-item-path="imported-item-block-4-item-1"><div class="list-item-paragraph">Determine the possible value(s) of $k$ such that&nbsp;$l_1$ and $l_2$ are skew.<span class="inline-marks">[3]</span></div></li></ol>
<p>Assume that&nbsp;$k = 4$ for the rest of this question.</p>
<ol class="i-list" start="3"><li data-list-item-id="imported-item-block-6-item-0" data-list-item-path="imported-item-block-6-item-0"><div class="list-item-paragraph">Points $A$ and $B$ are on $l_1$ and $l_2$ respectively such that&nbsp;$\overrightarrow{AB}$ is perpendicular to both lines. Show that&nbsp;$\left\vert{} \overrightarrow{AB} \right\vert{} = \sqrt{\frac{p}{2}}$, where&nbsp;$p$ is an integer to be determined.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="imported-item-block-6-item-1" data-list-item-path="imported-item-block-6-item-1"><div class="list-item-paragraph">Find exactly the sine of the acute angle between $l_2$ and $\pi_1$.<span class="inline-marks">[2]</span></div></li></ol>
<p>It is given further that&nbsp;$l_2$ lies on a third plane that is perpendicular to $l_1$, and $l_2$ intersects $\pi_1$ at point $P$.</p>
<ol class="i-list" start="5"><li data-list-item-id="imported-item-block-8-item-0" data-list-item-path="imported-item-block-8-item-0"><div class="list-item-paragraph">Deduce the shortest distance from $P$ to $l_1$.<span class="inline-marks">[2]</span></div></li></ol>

The planes $\pi_1$ and $\pi_2$, which meet in the line $l_1$, have equations

\[\pi_1 : \mathbf{r} \cdot \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} = 25 \quad \text{and} \quad \pi_2 : x + ky - 2z = -15,\]

Another line $l_2$ has equation $\mathbf{r} = \begin{pmatrix} -15 \\ 0 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}, \beta \in \mathbb{R}$.

Determine the position vector of the point on $l_1$ such that the coordinates of this point is independent of $k$. Hence find a vector equation of $l_1$.

Determine the possible value(s) of $k$ such that $l_1$ and $l_2$ are skew.

Assume that $k = 4$ for the rest of this question.

Points $A$ and $B$ are on $l_1$ and $l_2$ respectively such that $\overrightarrow{AB}$ is perpendicular to both lines. Show that $\left\vert{} \overrightarrow{AB} \right\vert{} = \sqrt{\frac{p}{2}}$, where $p$ is an integer to be determined.

Find exactly the sine of the acute angle between $l_2$ and $\pi_1$.

It is given further that $l_2$ lies on a third plane that is perpendicular to $l_1$, and $l_2$ intersects $\pi_1$ at point $P$.

Deduce the shortest distance from $P$ to $l_1$.

Common Perpendicular and Distance Between Skew Lines

Relationships Between Two Lines

Line of Intersection of Two Planes

Distance from a Point to a Line

2024 NJC P1 Q12

<p>The point $A$ has coordinates $(1, -1, 2)$ and the line $l$ has equation $\mathbf{r} = 2\mathbf{i} - 3\mathbf{j} + \lambda(\mathbf{i} - 2\mathbf{j} + 5\mathbf{k})$. The point $A$ and the line $l$ lies in the plane $p_1$.</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"><p>Find the position vector of the foot of perpendicular from $A$ to the line $l$.</p><span class="inline-marks">[3]</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">Show that the equation of $p_1$ is $2x + y = 1$.<span class="inline-marks">[2]</span></div></li></ol>
<p>The plane $p_2$ with equation $8x + ay + z = 4$ meets the plane $p_1$ at the line $L$ and the point $B(0, 1, 0)$ lies on $L$.</p>
<ol class="i-list" start="3"><li data-list-item-id="imported-item-block-4-item-0" data-list-item-path="imported-item-block-4-item-0"><div class="list-item-paragraph">Show that&nbsp;$L$ is parallel to $\mathbf{i} - 2\mathbf{j} + (2a - 8)\mathbf{k}$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="imported-item-block-4-item-1" data-list-item-path="imported-item-block-4-item-1"><div class="list-item-paragraph">Show that&nbsp;$a = 4$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="imported-item-block-4-item-2" data-list-item-path="imported-item-block-4-item-2"><div class="list-item-paragraph">Find the angle between $l$ and $p_2$.<span class="inline-marks">[2]</span></div></li></ol>
<p>Another point $C$ lies in the plane $p_1$ such that&nbsp;$BC$ is perpendicular to the line $L$.</p>
<ol class="i-list" start="6"><li data-list-item-id="imported-item-block-6-item-0" data-list-item-path="imported-item-block-6-item-0"><div class="list-item-paragraph">Show that&nbsp;$\overrightarrow{BC}$ is parallel to $\mathbf{k}$. Given that the distance of $C$ from $p_2$ is 5 units, find the possible coordinates of $C$.<span class="inline-marks">[4]</span></div></li></ol>

The point $A$ has coordinates $(1, -1, 2)$ and the line $l$ has equation $\mathbf{r} = 2\mathbf{i} - 3\mathbf{j} + \lambda(\mathbf{i} - 2\mathbf{j} + 5\mathbf{k})$. The point $A$ and the line $l$ lies in the plane $p_1$.

Show that the equation of $p_1$ is $2x + y = 1$.

The plane $p_2$ with equation $8x + ay + z = 4$ meets the plane $p_1$ at the line $L$ and the point $B(0, 1, 0)$ lies on $L$.

Show that $L$ is parallel to $\mathbf{i} - 2\mathbf{j} + (2a - 8)\mathbf{k}$.

Find the angle between $l$ and $p_2$.

Another point $C$ lies in the plane $p_1$ such that $BC$ is perpendicular to the line $L$.

Show that $\overrightarrow{BC}$ is parallel to $\mathbf{k}$. Given that the distance of $C$ from $p_2$ is 5 units, find the possible coordinates of $C$.

Perpendicular Projection and Reflection in a Line

Relationships Between a Point and a Plane

Relationships Between a Line and a Plane

<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 $90^\circ$.

2024 RI Promo Q2

<ol class="abc-list"><li data-list-item-id="item-a"><ol class="i-list"><li data-list-item-id="item-a-i"><div class="list-item-paragraph">Given that&nbsp;$\mathbf{p} \cdot \mathbf{q} = 0$, what can be deduced about the vectors $\mathbf{p}$ and $\mathbf{q}$?<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-a-ii"><div class="list-item-paragraph">With respect to the origin on a $xy$-plane, a fixed point $G$ has position vector $\mathbf{g}$ and a variable point $R$ has position vector $\mathbf{r}$. Describe the locus of $R$ if&nbsp;$\mathbf{r} \cdot (\mathbf{r} - \mathbf{g}) = 0$.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="item-b"><div class="list-item-paragraph">With respect to the origin, plane $p$ has an equation given by $\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) + \mu(2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k})$, $\lambda \in \mathbb{R}$, $\mu \in \mathbb{R}$. Points $Q$ and $R$ have position vectors $3\mathbf{i} + 9\mathbf{j} + 5\mathbf{k}$ and $4\mathbf{i} + 11\mathbf{j} + 7\mathbf{k}$ respectively. Find the length of the projection of $\overrightarrow{QR}$ onto $p$.<span class="inline-marks">[4]</span></div></li></ol>

2024 NJC Promo Q7

<p>The line $l_1$ has the vector equation $\mathbf{r} = \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ b \end{pmatrix}$, $\lambda \in \mathbb{R}$, where&nbsp;$a$ and $b$ are real constants.</p>
<p>The plane $p_1$ has cartesian equation $x + 2y + 4z = 4$.</p>
<p>The line $l_1$ is perpendicular to the line $l_2$ with cartesian equation $\frac{1-y}{2} = z - 3$, $x = 5$.</p>
<p>The line $l_1$ intersects the plane $p_1$ at the point with coordinates $(a, 1, 0)$.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">By first finding the vector equation of line $l_2$, show that&nbsp;$a = b = 2$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-paragraph">Verify that the point $A$ with coordinates $(2, 2, 2)$ lies on $l_1$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-iii"><div class="list-item-paragraph">Find the coordinates of the foot of perpendicular, $F$, from point $A$ to the plane $p_1$.<br>Hence find the shortest distance from point $A$ to the plane $p_1$.<span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-iv"><div class="list-item-paragraph">The plane $p_2$ contains the line $l_1$ and the point $F$. Find the vector equation of the plane $p_2$, giving your answer in the form $\mathbf{r} \cdot \mathbf{n} = d$, where&nbsp;$\mathbf{n}$ is a vector and $d$ is a real constant to be determined.<span class="inline-marks">[2]</span></div></li></ol>

The line $l_1$ has the vector equation $\mathbf{r} = \begin{pmatrix} a \\ 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 1 \\ b \end{pmatrix}$, $\lambda \in \mathbb{R}$, where $a$ and $b$ are real constants.

The plane $p_1$ has cartesian equation $x + 2y + 4z = 4$.

The line $l_1$ is perpendicular to the line $l_2$ with cartesian equation $\frac{1-y}{2} = z - 3$, $x = 5$.

The line $l_1$ intersects the plane $p_1$ at the point with coordinates $(a, 1, 0)$.

2024 ACJC Promo Q11

<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.

2006 JJC P1 Q12

<p>In the diagram below, $AB$ is a diameter of the circle with centre $O$. Point $C$ is on the circumference of the circle. Let&nbsp;$\overrightarrow{OB}=\mathbf{b}$ and $\overrightarrow{OC}=\mathbf{c}$.</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 data-list-item-id="item-1780161291959-45u9sbkrp" data-list-item-path="item-1780161291959-45u9sbkrp"><div class="list-item-paragraph">Find an expression for&nbsp;$\overrightarrow{CB}$ and for&nbsp;$\overrightarrow{AC}$ in terms of $\mathbf{b}$ and $\mathbf{c}$.</div></li><li data-list-item-id="item-1780161292512-sd8aqo5e7" data-list-item-path="item-1780161292512-sd8aqo5e7"><div class="list-item-paragraph">Hence prove that&nbsp;$ACB$ is a right angle.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="graph-block" data-graph-config="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" data-width="204" data-height="175" data-proportional style="width:204px;height:175px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:204px;--graph-aspect:204 / 175;"></div></div></div></div>

In the diagram below, $AB$ is a diameter of the circle with centre $O$. Point $C$ is on the circumference of the circle. Let $\overrightarrow{OB}=\mathbf{b}$ and $\overrightarrow{OC}=\mathbf{c}$.

Vector Subtraction

<p>The acute angle between two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ is $\alpha$ radians, and the vectors $\mathbf{b}$ and $2\mathbf{a}-\mathbf{b}$ are perpendicular. Find the value of $\alpha$ if&nbsp;$|\mathbf{b}| = \sqrt{2}|\mathbf{a}|$.<span class="inline-marks">[3]</span></p>

The acute angle between two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ is $\alpha$ radians, and the vectors $\mathbf{b}$ and $2\mathbf{a}-\mathbf{b}$ are perpendicular. Find the value of $\alpha$ if $|\mathbf{b}| = \sqrt{2}|\mathbf{a}|$.

Angles Between Vectors

2025 YIJC P2 Q1

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjkyNDA2NTg3MjAtZXIwcmkzeDBhM2wucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDk1MiwiZXhwIjoxNzkzNTA1MjUyfQ.7_z3gjcE7NqmWGjJBe549wMYw-qCYth6fgdvyJ8_OuE" 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.

Areas of Triangles and Parallelograms Using Vectors

2025 SAJC P1 Q11

<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 $p$.

2025 RVHS P1 Q6

<p>Two planes $p_1$ and $p_2$ have respective cartesian equations given by</p>
<p style="text-align: center">\[4x + y + z = 8 \text{ and}\hspace{0.5em} 4x + 3y - z = 0.\]</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Find the sine of the acute angle between $p_1$ and $p_2$ in the form $\frac{m}{\sqrt{n}}$ where&nbsp;$m$ and $n$ are positive integers to be determined.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Verify that the point $A$ with coordinates $(1,0,4)$ lies on $p_1$ and $p_2$. Hence, without the use of a calculator, find the vector equation of the line $l$ formed by the intersection of $p_1$ and $p_2$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">It is given that&nbsp;$B$ is a point on $p_1$ with coordinates $(1,3,1)$. Show that&nbsp;$AB$ is perpendicular to $l$ and hence use <strong>(a)</strong> to deduce exactly the shortest distance from $B$ to $p_2$.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the cartesian equation of the plane $p_3$ which contains $B$ and is perpendicular to both&nbsp;$p_1$ and $p_2$.</span><span class="inline-marks">[2]</span></div></li></ol>

Two planes $p_1$ and $p_2$ have respective cartesian equations given by

\[4x + y + z = 8 \text{ and } 4x + 3y - z = 0.\]

Find the sine of the acute angle between $p_1$ and $p_2$ in the form $\frac{m}{\sqrt{n}}$ where $m$ and $n$ are positive integers to be determined.

Verify that the point $A$ with coordinates $(1,0,4)$ lies on $p_1$ and $p_2$. Hence, without the use of a calculator, find the vector equation of the line $l$ formed by the intersection of $p_1$ and $p_2$.

Find the cartesian equation of the plane $p_3$ which contains $B$ and is perpendicular to both $p_1$ and $p_2$.

Angle Between Two Planes

2025 NYJC P2 Q5

<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>The points $A$ and $B$ have coordinates $(15, 3, 0)$ and $(5, 9, 5)$ respectively. Two lines $l_1$ and $l_2$, which are perpendicular to each other, have the following equations.</p>
<p style="margin-left: 7rem">$l_1 : \mathbf{r} = (15 - \lambda)\mathbf{i} + (3 + 2\lambda)\mathbf{j} + 4\lambda\mathbf{k},$<br>$l_2 : \mathbf{r} = (5 + 8\mu)\mathbf{i} + (9 - 2\mu)\mathbf{j} + (5 + m\mu)\mathbf{k},$</p>
<p>where&nbsp;$\lambda$ and $\mu$ are parameters and $m$ is a constant.</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 value of $m$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Given that&nbsp;$l_1$ and $l_2$ intersect, find the coordinates of the point of intersection $E$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Find a cartesian equation of the plane $\Pi$ which contains the points $A$, $B$ and $E$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">The point $D$ has coordinates $(-1, -3, 2)$. Find the position vector of the point $F$, the foot of perpendicular of $D$ to $\Pi$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-e" data-list-item-path="item-e"><div class="list-item-paragraph">Find the exact area of the circle that passes through $A$, $D$ and $F$.<span class="inline-marks">[2]</span></div></li></ol>

The points $A$ and $B$ have coordinates $(15, 3, 0)$ and $(5, 9, 5)$ respectively. Two lines $l_1$ and $l_2$, which are perpendicular to each other, have the following equations.

$l_1 : \mathbf{r} = (15 - \lambda)\mathbf{i} + (3 + 2\lambda)\mathbf{j} + 4\lambda\mathbf{k},$
$l_2 : \mathbf{r} = (5 + 8\mu)\mathbf{i} + (9 - 2\mu)\mathbf{j} + (5 + m\mu)\mathbf{k},$

where $\lambda$ and $\mu$ are parameters and $m$ is a constant.

Given that $l_1$ and $l_2$ intersect, find the coordinates of the point of intersection $E$.

Find a cartesian equation of the plane $\Pi$ which contains the points $A$, $B$ and $E$.

The point $D$ has coordinates $(-1, -3, 2)$. Find the position vector of the point $F$, the foot of perpendicular of $D$ to $\Pi$.

Find the exact area of the circle that passes through $A$, $D$ and $F$.

Intersection of Two Lines

2025 HCI P1 Q11

<p>Relative to the origin $O$, the position vectors of points $A$, $B$ and $C$ are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively. It is given that&nbsp;$\mathbf{a}$ is a unit vector and $\mathbf{b}$ is perpendicular to $\mathbf{b} - \mathbf{a}$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Show that&nbsp;$0 &lt; |\mathbf{b}| &lt; 1$.</span><span class="inline-marks">[3]</span></div></li></ol>
<p>It is given further that&nbsp;$|\mathbf{b}| = \frac{1}{2}$ and $C$ lies on $AB$ such that&nbsp;$AC : CB = 2 : 1$.</p>
<ol class="abc-list" start="2"><li><div class="list-item-content"><span class="list-item-text">Find the value of $|\mathbf{a} \cdot \mathbf{c}|$ and state the geometrical interpretation of $|\mathbf{a} \cdot \mathbf{c}|$.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the value of $\frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a} \times \mathbf{c}|}$.</span><span class="inline-marks">[2]</span></div></li></ol>

Relative to the origin $O$, the position vectors of points $A$, $B$ and $C$ are $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ respectively. It is given that $\mathbf{a}$ is a unit vector and $\mathbf{b}$ is perpendicular to $\mathbf{b} - \mathbf{a}$.

It is given further that $|\mathbf{b}| = \frac{1}{2}$ and $C$ lies on $AB$ such that $AC : CB = 2 : 1$.

Find the value of $|\mathbf{a} \cdot \mathbf{c}|$ and state the geometrical interpretation of $|\mathbf{a} \cdot \mathbf{c}|$.

Find the value of $\frac{|\mathbf{a} \times \mathbf{b}|}{|\mathbf{a} \times \mathbf{c}|}$.

2025 CJC P1 Q6

<p>There are three distinct points $A$, $B$ and $P$. The point $P$ lies on the circle with centre $O$ and diameter $AB$. Relative to the origin $O$, the position vectors of points $A$ and $P$ are $\mathbf{a}$ and $\mathbf{p}$ respectively.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">By expressing $\overrightarrow{AP}$ and $\overrightarrow{BP}$ in terms of $\mathbf{a}$ and $\mathbf{p}$, show that $\overrightarrow{AP}\cdot \overrightarrow{BP}=0$.</div><p style="text-align: right;">[3]</p></li></ol>
<p>The point $C$ with position vector $\mathbf{c}$ lies on $AB$ produced such that $AC:AB=\lambda :1$.</p>
<ol class="abc-list" start="2"><li><div class="list-item-paragraph">Find $\mathbf{c}$ in terms of $\lambda $ and $\mathbf{a}$.</div><p style="text-align: right;">[2]</p></li><li><div class="list-item-paragraph">It is given that $PC$ is a tangent to the circle and that angle $AOP={{120}^{{}^\circ }}$. Using a suitable scalar product, find the value of $\lambda $.</div><p style="text-align: right;">[4]</p></li></ol>

There are three distinct points $A$, $B$ and $P$. The point $P$ lies on the circle with centre $O$ and diameter $AB$. Relative to the origin $O$, the position vectors of points $A$ and $P$ are $\mathbf{a}$ and $\mathbf{p}$ respectively.

The point $C$ with position vector $\mathbf{c}$ lies on $AB$ produced such that $AC:AB=\lambda :1$.

Math Question Bank