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

<p>An adventure park is designing a new zipline that will span between two hilltops. The diagram below shows a vertical cross-section of the zipline setup. Points are defined relative to a main control centre at $(0, 0, 0)$. The ground is horizontal and is modelled as the $x$-$y$ plane.</p>
<div class="graph-block" data-graph-config="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" data-width="520" data-height="260" style="width:520px;height:260px;max-width:100%;display:block;margin:0 auto;"></div>
<p>The adventure park owner proposes to place the upper launch station at point $A(0, 75, 50)$, and the lower landing station at point $B(600, 25, 20)$. Points $C$ and $D$ are taken as the foot of perpendicular from points $B$ and $A$ onto the $x$-$y$ plane respectively.</p>
<ol class="abc-list"><li data-list-item-id="h2-2025-tjc-promo-q12-tjc-q12-item-block-3-item-0" data-list-item-path="h2-2025-tjc-promo-q12-tjc-q12-item-block-3-item-0"><div class="list-item-paragraph">Find a vector equation of line $AB$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-2025-tjc-promo-q12-tjc-q12-item-block-3-item-1" data-list-item-path="h2-2025-tjc-promo-q12-tjc-q12-item-block-3-item-1"><div class="list-item-paragraph">Write down the coordinates of $C$ and $D$. Hence or otherwise, find the slope angle of descent, $\beta$, for the proposed zipline.<span class="inline-marks">[3]</span></div></li></ol>
<p>A viewing platform is to be constructed at a scenic point in the valley, located at point $P(105, 30, 30)$.</p>
<ol class="abc-list" start="3"><li data-list-item-id="h2-2025-tjc-promo-q12-tjc-q12-item-block-5-item-0" data-list-item-path="h2-2025-tjc-promo-q12-tjc-q12-item-block-5-item-0"><div class="list-item-paragraph">Show that a vector equation of the plane $ABCD$ is $\mathbf{r}\cdot \begin{pmatrix} 1 \\ 12 \\ 0 \end{pmatrix}=900$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-tjc-promo-q12-tjc-q12-item-block-5-item-1" data-list-item-path="h2-2025-tjc-promo-q12-tjc-q12-item-block-5-item-1"><div class="list-item-paragraph">Find the exact coordinates of the foot of perpendicular from $P$ to the plane $ABCD$.<span class="inline-marks">[4]</span></div></li></ol>

An adventure park is designing a new zipline that will span between two hilltops. The diagram below shows a vertical cross-section of the zipline setup. Points are defined relative to a main control centre at $(0, 0, 0)$. The ground is horizontal and is modelled as the $x$-$y$ plane.

The adventure park owner proposes to place the upper launch station at point $A(0, 75, 50)$, and the lower landing station at point $B(600, 25, 20)$. Points $C$ and $D$ are taken as the foot of perpendicular from points $B$ and $A$ onto the $x$-$y$ plane respectively.

Write down the coordinates of $C$ and $D$. Hence or otherwise, find the slope angle of descent, $\beta$, for the proposed zipline.

A viewing platform is to be constructed at a scenic point in the valley, located at point $P(105, 30, 30)$.

Show that a vector equation of the plane $ABCD$ is $\mathbf{r}\cdot \begin{pmatrix} 1 \\ 12 \\ 0 \end{pmatrix}=900$.

Find the exact coordinates of the foot of perpendicular from $P$ to the plane $ABCD$.

Perpendicular Projection and Reflection in a Plane

Angle Between a Line and a Plane

Equations of Lines

Three-Dimensional Geometry

Equations of Planes

2025 TJC Promo Q12

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

Vector (Cross) Product

Areas of Triangles and Parallelograms Using Vectors

Scalar (Dot) Product

2025 SAJC Promo Q8

<p>With reference to the origin $O$, the point $A$ has position vector $-\mathbf{i}-3\mathbf{j}+3\mathbf{k}$. The plane $\pi$ and the line $l$ have equations \[\mathbf{r}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\begin{pmatrix}-1\\3\\2\end{pmatrix}+\mu\begin{pmatrix}2\\3\\-4\end{pmatrix}\text{ and}\hspace{0.5em}\mathbf{r}=\begin{pmatrix}2\\1\\1\end{pmatrix}+\beta\begin{pmatrix}3\\t\\-2\end{pmatrix},\] respectively, where&nbsp;$\lambda$, $\mu$ and $\beta$ are parameters and $t$ is a real value.</p>
<ol class="abc-list"><li data-list-item-id="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-0" data-list-item-path="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-0"><div class="list-item-paragraph">It is given that&nbsp;$l$ and $\pi$ intersect at a point.</div><ol class="i-list"><li data-list-item-id="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-0-sub-0" data-list-item-path="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-0/h2-2025-ri-promo-q10-ri-q10-item-block-3-item-0-sub-0"><div class="list-item-paragraph">Find the range of values of $t$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-0-sub-1" data-list-item-path="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-0/h2-2025-ri-promo-q10-ri-q10-item-block-3-item-0-sub-1"><div class="list-item-paragraph">Find, in terms of $t$, the coordinates of the point of intersection, $B$, of $l$ and $\pi$.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1" data-list-item-path="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1"><div class="list-item-paragraph">It is given that&nbsp;$A$ lies on $l$.</div><ol class="i-list"><li data-list-item-id="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1-sub-0" data-list-item-path="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1/h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1-sub-0"><div class="list-item-paragraph">Show that&nbsp;$t=4$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1-sub-1" data-list-item-path="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1/h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1-sub-1"><div class="list-item-paragraph">Find the position vector of the foot of perpendicular, $F$, of $A$ onto $\pi$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1-sub-2" data-list-item-path="h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1/h2-2025-ri-promo-q10-ri-q10-item-block-3-item-1-sub-2"><div class="list-item-paragraph">Find the cartesian equation of another line $m$ through $A$, that lies on the plane containing $A$, $B$ and $F$, and $m$ makes the same angle with $\pi$ as&nbsp;$l$ makes with $\pi$.<span class="inline-marks">[4]</span></div></li></ol></li></ol>

With reference to the origin $O$, the point $A$ has position vector $-\mathbf{i}-3\mathbf{j}+3\mathbf{k}$. The plane $\pi$ and the line $l$ have equations \[\mathbf{r}=\begin{pmatrix}4\\1\\1\end{pmatrix}+\lambda\begin{pmatrix}-1\\3\\2\end{pmatrix}+\mu\begin{pmatrix}2\\3\\-4\end{pmatrix}\text{ and }\mathbf{r}=\begin{pmatrix}2\\1\\1\end{pmatrix}+\beta\begin{pmatrix}3\\t\\-2\end{pmatrix},\] respectively, where $\lambda$, $\mu$ and $\beta$ are parameters and $t$ is a real value.

It is given that $l$ and $\pi$ intersect at a point.

Relationships Between a Line and a Plane

Intersection of a Line and a Plane

2025 RI Promo Q10

<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

Perpendicular Vectors Using the Scalar Product

Geometrical Interpretation of Vector Equations

2025 RI Promo Q6

<p>The line $l$ passes through the points $A(2,-1, 3)$ and $B(4, 1, 5)$. The plane $p_1$ has equation $2x-y+2z=0$.</p>
<ol class="i-list"><li data-list-item-id="h2-2025-mi-promo-q10-mi-q10-item-block-1-item-0" data-list-item-path="h2-2025-mi-promo-q10-mi-q10-item-block-1-item-0"><div class="list-item-paragraph">Find the vector equation of the line $l$.<span class="inline-marks">[2]</span></div></li></ol>
<p>The plane $p_2$ contains $l$ and the point $C$ with coordinates $(3, 6, 0)$.</p>
<ol class="i-list" start="2"><li data-list-item-id="h2-2025-mi-promo-q10-mi-q10-item-block-3-item-0" data-list-item-path="h2-2025-mi-promo-q10-mi-q10-item-block-3-item-0"><div class="list-item-paragraph"><p>Show that a vector perpendicular to $p_2$ is $\begin{pmatrix}-5\\2\\3\end{pmatrix}$.</p><p>Hence find the cartesian equation of $p_2$.</p><span class="inline-marks">[3]</span></div></li><li data-list-item-id="h2-2025-mi-promo-q10-mi-q10-item-block-3-item-1" data-list-item-path="h2-2025-mi-promo-q10-mi-q10-item-block-3-item-1"><div class="list-item-paragraph">Find the vector equation of the line of intersection between plane $p_1$ and plane $p_2$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-2025-mi-promo-q10-mi-q10-item-block-3-item-2" data-list-item-path="h2-2025-mi-promo-q10-mi-q10-item-block-3-item-2"><div class="list-item-paragraph">Find the acute angle between the planes $p_1$ and $p_2$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="h2-2025-mi-promo-q10-mi-q10-item-block-3-item-3" data-list-item-path="h2-2025-mi-promo-q10-mi-q10-item-block-3-item-3"><div class="list-item-paragraph">Find the position vector of the point $F$, the foot of perpendicular from point $B$ to the plane $p_1$.<span class="inline-marks">[4]</span></div></li></ol>

The line $l$ passes through the points $A(2,-1, 3)$ and $B(4, 1, 5)$. The plane $p_1$ has equation $2x-y+2z=0$.

Find the vector equation of the line $l$.

The plane $p_2$ contains $l$ and the point $C$ with coordinates $(3, 6, 0)$.

Find the vector equation of the line of intersection between plane $p_1$ and plane $p_2$.

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

Find the position vector of the point $F$, the foot of perpendicular from point $B$ to the plane $p_1$.

Angle Between Two Planes

Line of Intersection of Two Planes

2025 MI PU1 Promo Q10

<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/1783577066686-nwkpkpnl7cj.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODM1NzcwNjY2ODYtbndrcGtwbmw3Y2oucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDg2MiwiZXhwIjoxNzkzNTA1MTYyfQ.Vm-AjTroOKo8rN9VcNol_R8-eWOM51vrcLnvVsAbV58" style="max-width: min(180px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>The origin $O$ and the points $A$, $B$ and $C$ lie in the same plane, where&nbsp;$\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$ and $\overrightarrow{OC} = \mathbf{c}$. It is given that&nbsp;$\angle AOC = \angle BOC = \theta$, where&nbsp;$\theta$ is an acute angle.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that&nbsp;$\mathbf{c} \cdot \mathbf{\hat{a}} = \mathbf{c} \cdot \mathbf{\hat{b}}$ where&nbsp;$\mathbf{\hat{a}}$ and $\mathbf{\hat{b}}$ are unit vectors in the directions of vectors $\mathbf{a}$ and $\mathbf{b}$ respectively.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">If&nbsp;$\mathbf{c}$ can be written as&nbsp;$m\mathbf{\hat{a}} + n\mathbf{\hat{b}}$, where&nbsp;$m$ and $n$ are constants, use the result from <strong>(a)</strong> to show that&nbsp;$m = n$.<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">Write down the equation of the line passing through the points $A$ and $B$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-d" data-list-item-path="item-d"><div class="list-item-paragraph">Given that&nbsp;$|\mathbf{a}| = 3$, $|\mathbf{b}| = 2$ and $m = n$, show that the position vector of the point of intersection of the line passing through $A$ and $B$ and the line passing through $O$ and $C$ is $t\left(\mathbf{\hat{a}} + \mathbf{\hat{b}}\right)$, where&nbsp;$t$ is a constant to be determined.<span class="inline-marks">[3]</span></div></li></ol>

The origin $O$ and the points $A$, $B$ and $C$ lie in the same plane, where $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OB} = \mathbf{b}$ and $\overrightarrow{OC} = \mathbf{c}$. It is given that $\angle AOC = \angle BOC = \theta$, where $\theta$ is an acute angle.

Intersection of Two Lines

2024 ACJC P2 Q4

<p>The line $l$ passes through $A(1, 2, -1)$ and is parallel to $3\mathbf{i} + 3\mathbf{k}$.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">State the vector equation of $l$.<span class="inline-marks">[1]</span></div></li></ol>
<p>The plane $p$ has equation</p>
<p style="text-align: center">\[\mathbf{r} \cdot \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = 13.\]</p>
<ol class="i-list" start="2"><li data-list-item-id="item-ii"><div class="list-item-paragraph">Find the acute angle between $l$ and $p$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iii"><div class="list-item-paragraph">Find the point of intersection of $l$ and $p$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iv"><div class="list-item-paragraph">Find the vector equation of the reflection of $l$ in $p$.<span class="inline-marks">[6]</span></div></li></ol>

The line $l$ passes through $A(1, 2, -1)$ and is parallel to $3\mathbf{i} + 3\mathbf{k}$.

\[\mathbf{r} \cdot \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} = 13.\]

<p>The line $l$ passes through $A(3, 1, 4)$ and is parallel to $2\mathbf{i} + \mathbf{j} + \mathbf{k}$.</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">State the vector equation of $l$.<span class="inline-marks">[1]</span></div></li></ol>
<p>The plane $p$ has equation</p>
<p style="text-align: center">\[\mathbf{r} \cdot \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix} = 7.\]</p>
<ol class="i-list" start="2"><li data-list-item-id="item-ii"><div class="list-item-paragraph">Find the acute angle between $l$ and $p$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iii"><div class="list-item-paragraph">Find the point of intersection of $l$ and $p$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iv"><div class="list-item-paragraph">Find the vector equation of the reflection of $l$ in $p$.<span class="inline-marks">[6]</span></div></li></ol>

The line $l$ passes through $A(3, 1, 4)$ and is parallel to $2\mathbf{i} + \mathbf{j} + \mathbf{k}$.

\[\mathbf{r} \cdot \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix} = 7.\]

<p>The points $A$ and $B$ have coordinates $(2, 1, -1)$ and $(5, -1, 3)$ respectively.</p>
<ol class="abc-list"><li data-list-item-id="item-1"><div class="list-item-paragraph">Find a vector equation of the line $AB$.<span class="inline-marks">[1]</span></div></li></ol>
<p>The point $C$ has coordinates $(10, -2, 6)$.</p>
<ol class="abc-list" start="2"><li data-list-item-id="item-2"><div class="list-item-paragraph">Find the position vector of the foot of perpendicular from $C$ to the line $AB$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-3"><div class="list-item-paragraph">Find a vector equation of the line which is a reflection of the line $AC$ in the line $AB$.<span class="inline-marks">[3]</span></div></li></ol>

The points $A$ and $B$ have coordinates $(2, 1, -1)$ and $(5, -1, 3)$ respectively.

The point $C$ has coordinates $(10, -2, 6)$.

Perpendicular Projection and Reflection in a Line

<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 $(a, 1, 0)$.

Relationships Between a Point and a Plane

Distance from a Point to a Plane

2024 ACJC Promo Q11

<p>The line $l_1$ has equation $x - 1 = 2 - y = -z$.</p>
<p>The line $l_2$ passes through the point $A(-2, 1, 7)$ and is parallel to the vector $-3\mathbf{i} + \mathbf{j} + 5\mathbf{k}$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Write down vector equations for&nbsp;$l_1$ and $l_2$.</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Show that the lines $l_1$ and $l_2$ intersect and find the coordinates of $E$, the point of intersection of $l_1$ and $l_2$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">The acute angle between the lines $l_1$ and $l_2$ is denoted by $\theta$. By finding $\cos \theta$, show that \[\sin \theta = \frac{2\sqrt{2}}{\sqrt{35}}.\] Hence find the shortest distance from $A$ to the line $l_1$, leaving your answer in exact form.</span><span class="inline-marks">[5]</span></div></li></ol>

The line $l_1$ has equation $x - 1 = 2 - y = -z$.

The line $l_2$ passes through the point $A(-2, 1, 7)$ and is parallel to the vector $-3\mathbf{i} + \mathbf{j} + 5\mathbf{k}$.

Write down vector equations for $l_1$ and $l_2$.

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

The acute angle between the lines $l_1$ and $l_2$ is denoted by $\theta$. By finding $\cos \theta$, show that \[\sin \theta = \frac{2\sqrt{2}}{\sqrt{35}}.\] Hence find the shortest distance from $A$ to the line $l_1$, leaving your answer in exact form.

Angles Between Two Lines

Distance from a Point to a Line

<p>Physicists are investigating the reflective property of a particular reflective surface. The diagram below shows the set-up of a particular experiment, where a laser emitting device was placed at the point with coordinates $(1, 2, 3)$. A laser beam was emitted in the direction parallel to $\mathbf{i} + \mathbf{k}$. The path of the emitted laser beam and its reflected path make the same angle $\theta$ with the reflective surface. The plane containing these two paths is perpendicular to the reflective surface.</p>
<p>Write down the vector equation of the path of the emitted laser beam.<span class="inline-marks">[1]</span></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/1774500717235-6nupuw71o97.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzQ1MDA3MTcyMzUtNm51cHV3NzFvOTcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDg2MiwiZXhwIjoxNzkzNTA1MTYyfQ.V_Bz6-DqWBPZzE48kDaRMcRXFUP9jytbHTa4P_MIgwo" style="max-width: min(400px, 100%); height: auto; display: block;" alt="Diagram of laser reflection" class="aligncenter custom-width" /></figure></div>
<p>It is known that the reflective surface has equation $x + y + z = 4$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find $\theta$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Show that the laser beam meets the reflective surface at the point $(0, 2, 2)$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the vector equation of the path of the reflected laser beam.</span><span class="inline-marks">[5]</span></div></li></ol>

Physicists are investigating the reflective property of a particular reflective surface. The diagram below shows the set-up of a particular experiment, where a laser emitting device was placed at the point with coordinates $(1, 2, 3)$. A laser beam was emitted in the direction parallel to $\mathbf{i} + \mathbf{k}$. The path of the emitted laser beam and its reflected path make the same angle $\theta$ with the reflective surface. The plane containing these two paths is perpendicular to the reflective surface.

Write down the vector equation of the path of the emitted laser beam.

It is known that the reflective surface has equation $x + y + z = 4$.

Show that the laser beam meets the reflective surface at the point $(0, 2, 2)$.

Find the vector equation of the path of the reflected laser beam.

2017 VJC P1 Q11

<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/1775962953671-l5kobtsrm5n.jpg?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzU5NjI5NTM2NzEtbDVrb2J0c3JtNW4uanBnIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDg2MiwiZXhwIjoxNzkzNTA1MTYyfQ.IpKoz62ZWEcTMf8Aqy5MHcswwjVjqo7XE9OTZYg9D98" style="max-width: min(270px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>In the diagram, $O$ is the centre of the rectangular base $ABCD$ of a right pyramid with vertex $V$. Perpendicular unit vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ are parallel to $AB$, $BC$ and $OV$ respectively. The length of $AB$, $BC$ and $OV$ are $12\text{ cm}$, $6\text{ cm}$ and $6\text{ cm}$ respectively. A line $l$ has Cartesian equation $\frac{-x-4}{10} = y+2 = \frac{t-z}{-2}.$</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find a vector equation of $AV$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">If the line $l$ intersects $AV$ at $M$, find the position vector of $M$ and the value of $t$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the acute angle between $AV$ and the line $l$. Hence find the perpendicular distance from $A$ to the line $l$.</span></div></li></ol>

In the diagram, $O$ is the centre of the rectangular base $ABCD$ of a right pyramid with vertex $V$. Perpendicular unit vectors $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ are parallel to $AB$, $BC$ and $OV$ respectively. The length of $AB$, $BC$ and $OV$ are $12\text{ cm}$, $6\text{ cm}$ and $6\text{ cm}$ respectively. A line $l$ has Cartesian equation $\frac{-x-4}{10} = y+2 = \frac{t-z}{-2}.$

If the line $l$ intersects $AV$ at $M$, find the position vector of $M$ and the value of $t$.

Find the acute angle between $AV$ and the line $l$. Hence find the perpendicular distance from $A$ to the line $l$.

2006 HCI P1 Q14

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

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

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

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

Ratio Theorem and Section Formula

N2015 P1 Q7

<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.

Relationships Between Two Lines

Vectors in a Parallelogram

2006 JJC P1 Q12

<p>Find the Parametric and Cartesian equations of line which passes through the points $A\left( \begin{matrix}
   -2  \\
   1  \\
   3  \\
\end{matrix} \right)$ and $B\left( \begin{matrix}
   2  \\
   0  \\
   8  \\
\end{matrix} \right)$.<br></p>

Find the Parametric and Cartesian equations of line which passes through the points $A\left( \begin{matrix}
   -2  \\
   1  \\
   3  \\
\end{matrix} \right)$ and $B\left( \begin{matrix}
   2  \\
   0  \\
   8  \\
\end{matrix} \right)$.


<p></p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">The non-zero vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are such that&nbsp;$\mathbf{a} \times \mathbf{b} = \mathbf{c} \times \mathbf{a}$. Given that&nbsp;$\mathbf{b} \neq -\mathbf{c}$, find a linear relationship between $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">The variable position vector $\mathbf{v}$ satisfies the equation $\mathbf{v} \times (\mathbf{i} - 3\mathbf{k}) = 2\mathbf{j}$.</div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find the set of position vectors $\mathbf{v}$.</span><span class="inline-marks">[3]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Describe the set of points represented by position vectors $\mathbf{v}$ geometrically.</span><span class="inline-marks">[2]</span></div></li></ol></li></ol>

The non-zero vectors $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$ are such that $\mathbf{a} \times \mathbf{b} = \mathbf{c} \times \mathbf{a}$. Given that $\mathbf{b} \neq -\mathbf{c}$, find a linear relationship between $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{c}$.

2025 Specimen Paper P1 Q6

<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>Relative to the origin $O$, point $A$ has position vector given by $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$. The line $l_1$ passes through the point $A$ and is parallel to $\begin{pmatrix} t \\ 3 - t^2 \\ 1 \end{pmatrix}$. The plane $p$ has equation $\mathbf{r} = \lambda \begin{pmatrix} 0 \\ 5 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 5 \\ 0 \\ t \end{pmatrix}, \lambda, \mu \in \mathbb{R}$ where&nbsp;$t$ is a real constant. It is known that&nbsp;$l_1$ and $p$ are parallel and $l_1$ is not on $p$.</p>
<ol class="abc-list"><li data-list-item-id="item-1768711344529-8nihmues6" data-list-item-path="item-1768711344529-8nihmues6"><div class="list-item-paragraph">Show that&nbsp;$t = -1$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-1768711345279-s547r63ey" data-list-item-path="item-1768711345279-s547r63ey"><div class="list-item-paragraph">The line $l_2$ passes through the origin and point $A$. Find the acute angle between $l_1$ and $l_2$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1768711345923-79fmat0do" data-list-item-path="item-1768711345923-79fmat0do"><div class="list-item-paragraph">The vector $\mathbf{n}$ is a unit vector normal to $p$. State the geometrical meaning of $|\mathbf{a} \cdot \mathbf{n}|$ and find the exact value of $|\mathbf{a} \cdot \mathbf{n}|$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-1768711346929-qm8zctapw" data-list-item-path="item-1768711346929-qm8zctapw"><div class="list-item-paragraph">Find a vector equation of the line of reflection of $l_2$ in $p$.<span class="inline-marks">[4]</span></div></li></ol>

Relative to the origin $O$, point $A$ has position vector given by $\mathbf{a} = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$. The line $l_1$ passes through the point $A$ and is parallel to $\begin{pmatrix} t \\ 3 - t^2 \\ 1 \end{pmatrix}$. The plane $p$ has equation $\mathbf{r} = \lambda \begin{pmatrix} 0 \\ 5 \\ 2 \end{pmatrix} + \mu \begin{pmatrix} 5 \\ 0 \\ t \end{pmatrix}, \lambda, \mu \in \mathbb{R}$ where $t$ is a real constant. It is known that $l_1$ and $p$ are parallel and $l_1$ is not on $p$.

2025 EJC P2 Q4

<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/1768703533158-y2y9mw7e2d.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njg3MDM1MzMxNTgteTJ5OW13N2UyZC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0ODYyLCJleHAiOjE3OTM1MDUxNjJ9.44Myp_bMLYhQwpAvkAd7BDhd36ZBu_J5G8UZTw8WX04" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>Relative to the origin $O$, the points $A$, $B$, and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $3\mathbf{a}+\mathbf{b}$ respectively. The point $D$ lies on $AB$ such that&nbsp;$AD=kAB$, where&nbsp;$0&lt;k&lt;1$.</p>
<ol class="abc-list"><li data-list-item-id="item-6-a"><div class="list-item-content"><span class="list-item-text">Find a vector equation of the line $OD$ in terms of $k$, $\mathbf{a}$ and $\mathbf{b}$.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-6-b"><div class="list-item-content"><span class="list-item-text">The point $E$ is the midpoint of $BC$. Find the value of $k$ if&nbsp;$O$, $D$ and $E$ are collinear.</span><span class="inline-marks">[4]</span></div></li></ol>
<p>It is given that&nbsp;$|\mathbf{a}|=1$, $|\mathbf{b}|=2$ and $|3\mathbf{a}-2\mathbf{b}|=\sqrt{31}$.</p>
<ol class="abc-list" start="3"><li data-list-item-id="item-6-c"><div class="list-item-content"><span class="list-item-text">By considering the scalar product $(3\mathbf{a}-2\mathbf{b}) \cdot (3\mathbf{a}-2\mathbf{b})$, find the numerical value of $\mathbf{a} \cdot \mathbf{b}$ and hence determine the angle between $\mathbf{a}$ and $\mathbf{b}$.</span><span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-6-d"><div class="list-item-content"><span class="list-item-text">Give a geometrical interpretation of $|\mathbf{a} \cdot \mathbf{b}|$.</span><span class="inline-marks">[1]</span></div></li></ol>

Relative to the origin $O$, the points $A$, $B$, and $C$ have position vectors $\mathbf{a}$, $\mathbf{b}$ and $3\mathbf{a}+\mathbf{b}$ respectively. The point $D$ lies on $AB$ such that $AD=kAB$, where $0<k<1$.

Find a vector equation of the line $OD$ in terms of $k$, $\mathbf{a}$ and $\mathbf{b}$.

The point $E$ is the midpoint of $BC$. Find the value of $k$ if $O$, $D$ and $E$ are collinear.

It is given that $|\mathbf{a}|=1$, $|\mathbf{b}|=2$ and $|3\mathbf{a}-2\mathbf{b}|=\sqrt{31}$.

By considering the scalar product $(3\mathbf{a}-2\mathbf{b}) \cdot (3\mathbf{a}-2\mathbf{b})$, find the numerical value of $\mathbf{a} \cdot \mathbf{b}$ and hence determine the angle between $\mathbf{a}$ and $\mathbf{b}$.

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

Math Question Bank