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PSLE Foundation Mathematics

<p>When subtracting two vectors $\overrightarrow{AB}$ and $\overrightarrow{CB}$:</p>
<p style="text-align: center">$\overrightarrow{AB} - \overrightarrow{CB} = \overrightarrow{AC}$</p>
<p>For any two column vectors, $\mathbf{a} = \begin{pmatrix} p \\ q \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} r \\ s \end{pmatrix}$, then</p>
<p style="margin-left: 12.5rem">$\!\!\!\!$ $\mathbf{a} - \mathbf{b} = \begin{pmatrix} p - r \\ q - s \end{pmatrix}$</p>

When subtracting two vectors $\overrightarrow{AB}$ and $\overrightarrow{CB}$:

$\overrightarrow{AB} - \overrightarrow{CB} = \overrightarrow{AC}$

For any two column vectors, $\mathbf{a} = \begin{pmatrix} p \\ q \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} r \\ s \end{pmatrix}$, then

$\!\!\!\!$ $\mathbf{a} - \mathbf{b} = \begin{pmatrix} p - r \\ q - s \end{pmatrix}$

Vector Subtraction

Vectors

<p>To construct a special tentage, $3$ steel poles $OR$, $AQ$ and $BP$ are secured to the horizontal ground such that each of the poles are perpendicular to the ground. The point $O$ on the ground is taken as the origin and the unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ are parallel to $OA$, $OB$ and $OR$ respectively, with units in metres. It is given that&nbsp;$OR=12$, $AQ=9$, $BP=6$, $OA=20$ and $OB=15$ (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/1771903221872-mq6u65oef5.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzE5MDMyMjE4NzItbXE2dTY1b2VmNS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0OTEwLCJleHAiOjE3OTM1MDUyMTB9.2eJxqS8UCGb8LrGl2mLnwQXs9Uq2EIKmo3rpqWzBips" style="max-width: min(350px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find $\vec{PR}$ and $\vec{PQ}$.<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"><p>The canvas roof $PQR$, which is of negligible thickness, is part of the plane $p$. Using your answers in part (i), show that the equation of $p$ is given by ${\mathbf{r} \cdot \begin{pmatrix} 3 \\ 8 \\ 20 \end{pmatrix} = 240.}$</p><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find the acute angle between the canvas roof and the plane $OAB$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">A cable is to be installed close to the tentage. This cable is part of the line $L$ with vector equation 	$\mathbf{r}=\left( \begin{matrix}
   \begin{matrix}
   -2  \\
   8  \\
   12  \\
\end{matrix}  \\
\end{matrix} \right)+\lambda \left( \begin{matrix}
   \begin{matrix}
   20  \\
   13  \\
   -4  \\
\end{matrix}  \\
\end{matrix} \right)$, where&nbsp;$\lambda \in \mathbb{R}$. Show that this cable is not parallel to the edge $PR$ of the roof and does not meet 	the edge $PR$. <span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-v" data-list-item-path="item-v"><div class="list-item-paragraph">A lamp is hung vertically downwards from the canvas roof using a cable such that it is at the point $M$ with coordinates 	$\left( \frac{20}{3},5,8 \right)$. To secure the lamp, a second cable is connected from the lamp to the point $N$ on the roof. It is assumed 	that the cables are laid in straight lines and have negligible thickness. For the shortest length to be used for the second 	cable, show that&nbsp;$\overrightarrow{ON}=\left( \begin{matrix}
   \begin{matrix}
   s  \\
   \frac{2525}{473}  \\
   t  \\
\end{matrix}  \\
\end{matrix} \right)$, where&nbsp;$s$ and $t$ are constants to be found. <br><span class="inline-marks">[3]</span></div></li></ol>

To construct a special tentage, $3$ steel poles $OR$, $AQ$ and $BP$ are secured to the horizontal ground such that each of the poles are perpendicular to the ground. The point $O$ on the ground is taken as the origin and the unit vectors $\mathbf{i}$, $\mathbf{j}$ and $\mathbf{k}$ are parallel to $OA$, $OB$ and $OR$ respectively, with units in metres. It is given that $OR=12$, $AQ=9$, $BP=6$, $OA=20$ and $OB=15$ (see diagram).

Find the acute angle between the canvas roof and the plane $OAB$.

Perpendicular Projection and Reflection in a Plane

Three-Dimensional Geometry

Equations of Planes

Angle Between Two Planes

Relationships Between Two Lines

2024 MI PRU1 P1 Q9

<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 $\overrightarrow{OC}=\mathbf{c}$.

Perpendicular Vectors Using the Scalar Product

<p>In the diagram, $OA$ is parallel to $CB$ and $SC$ is parallel to $AB$. The position vectors of $A$ and $B$ relative to $O$ are $4\mathbf{a}$ and $5\mathbf{b}$ respectively. It is given that&nbsp;$OS : SB = 2 : 3$ and $\overrightarrow{AT} = \frac{1}{3}\overrightarrow{TB}$.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Find, in terms of $\mathbf{a}$ and $\mathbf{b}$, the vectors</div><ol class="i-list"><li><div class="list-item-paragraph">$\overrightarrow{OS}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{AS}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{OT}$.</div></li></ol></li><li><div class="list-item-paragraph">$U$ is a point on $OT$ such that&nbsp;$OU : UT = 8 : 3$.</div><ol class="i-list"><li><div class="list-item-paragraph">Express $\overrightarrow{AU}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.</div></li><li><div class="list-item-paragraph">Hence write down two facts about $A$, $U$ and $S$.</div></li></ol></li><li><div class="list-item-paragraph">Prove that&nbsp;$\Delta SBC$ is similar to $\Delta BOA$.</div></li><li><div class="list-item-paragraph">Find</div><ol class="i-list"><li><div class="list-item-paragraph">$\frac{\text{Area of}\hspace{0.5em} \Delta SBC}{\text{Area of}\hspace{0.5em} \Delta BOA}$,</div></li><li><div class="list-item-paragraph">$\frac{\text{Area of}\hspace{0.5em} \Delta OTA}{\text{Area of}\hspace{0.5em} \Delta OBA}$.</div></li></ol></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768041714968-469zggyiadd.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwNDE3MTQ5NjgtNDY5emdneWlhZGQucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.XVjyW_PiCBbqqFjMsIYVInIMT3H1MzbAdJ82Z9417Xg" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, $OA$ is parallel to $CB$ and $SC$ is parallel to $AB$. The position vectors of $A$ and $B$ relative to $O$ are $4\mathbf{a}$ and $5\mathbf{b}$ respectively. It is given that $OS : SB = 2 : 3$ and $\overrightarrow{AT} = \frac{1}{3}\overrightarrow{TB}$.

Collinearity Using Vectors

Area Ratios Using Vectors

Ratio Theorem and Section Formula

Vectors: Properties of Parallelism and Intersections

<p>In the triangle $OAB$, $M$ is the midpoint of $OA$. $N$ is a point on $OB$ such that&nbsp;$ON : NB = 2 : 1$. $MN$ is produced to $P$ so that&nbsp;$MN : NP = 1 : 2$.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>Given&nbsp;$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$.</p><ol class="abc-list"><li><div class="list-item-paragraph">Express, in terms of $\mathbf{a}$ and/or $\mathbf{b}$,</div><ol class="i-list"><li><div class="list-item-paragraph">$\overrightarrow{NB}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{MN}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{NP}$.</div></li></ol></li><li><div class="list-item-paragraph">Express $\overrightarrow{AB}$ and $\overrightarrow{BP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.</div></li><li><div class="list-item-paragraph">Write down two facts about $\overrightarrow{AB}$ and $\overrightarrow{BP}$.</div></li><li><div class="list-item-paragraph">Find $\frac{\text{area of triangle}\hspace{0.5em} PMB}{\text{area of triangle}\hspace{0.5em} PMA}$.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768039433891-5h4lxsehlwo.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMzk0MzM4OTEtNWg0bHhzZWhsd28ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.kYvz6IVGOrgfpj8BujqZAXQyoBO3hlMoQbv9QbQpmeo" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the triangle $OAB$, $M$ is the midpoint of $OA$. $N$ is a point on $OB$ such that $ON : NB = 2 : 1$. $MN$ is produced to $P$ so that $MN : NP = 1 : 2$.

Parallel Vectors

Properties of Parallelograms and Trapezia in Vector Form

<p>In the diagram, $OACB$ is a trapezium where&nbsp;$AC$ is parallel to $OB$. The line $OA$ is produced to the point $D$ such that&nbsp;$\frac{OA}{AD} = \frac{1}{2}$.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Given that&nbsp;$\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OB} = \mathbf{b}$, express, as simply as possible, in terms of $\mathbf{a}$ and/or $\mathbf{b}$,</div><ol class="i-list"><li><div class="list-item-paragraph">$\overrightarrow{BD}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{OC}$.</div></li></ol></li><li><div class="list-item-paragraph">Given that&nbsp;$\overrightarrow{OE} = 3\mathbf{a} + 2\mathbf{b}$,</div><ol class="i-list"><li><div class="list-item-paragraph">state the name of the quadrilateral $ODEB$,</div></li><li><div class="list-item-paragraph">explain why $O$, $C$ and $E$ lie in a straight line.</div></li></ol></li><li><div class="list-item-paragraph">Find, giving your answers as fractions in the simplest form,</div><ol class="i-list"><li><div class="list-item-paragraph">$\frac{\text{area of}\hspace{0.5em} \Delta ADC}{\text{area of}\hspace{0.5em} \Delta ODB}$,</div></li><li><div class="list-item-paragraph">$\frac{\text{area of}\hspace{0.5em} \Delta ADC}{\text{area of quadrilateral}\hspace{0.5em} ODEB}$.</div></li></ol></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768038939753-seoov8toh7.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMzg5Mzk3NTMtc2Vvb3Y4dG9oNy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0OTEwLCJleHAiOjE3OTM1MDUyMTB9.c2dMvcPP0xlu_Fgcp4oQMaJ9oHpB8nFtmE1aW0GmqZc" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

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

Vectors in a Parallelogram

Vector Geometry of Parallelograms and Trapezia

<p>The points $A$, $B$ and $C$ are vertices of a triangle. The point $A$ is $(-1,3)$ and the gradient of the line $AB$ is $2$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Find the equation of the line $AB$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">The equation of the line $BC$ is $2x+y=17$. Find the coordinates of $B$.</span></div></li><li><div class="list-item-paragraph">It is given that&nbsp;$\overrightarrow{AE} = \begin{pmatrix} 4 \\ 12 \end{pmatrix}$. Find the coordinates of $E$.</div></li><li><div class="list-item-content"><span class="list-item-text">Given also that&nbsp;$D$ is $(3,7)$, express as a column vector:</span></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">$\overrightarrow{DE}$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">$\overrightarrow{DB}$.</span></div></li></ol></li><li><div class="list-item-content"><span class="list-item-text">What two facts can be deduced about $D$, $E$ and $B$.</span></div></li></ol>

The points $A$, $B$ and $C$ are vertices of a triangle. The point $A$ is $(-1,3)$ and the gradient of the line $AB$ is $2$.

The equation of the line $BC$ is $2x+y=17$. Find the coordinates of $B$.

Given also that $D$ is $(3,7)$, express as a column vector:

What two facts can be deduced about $D$, $E$ and $B$.

Position Vectors

Coordinate Geometry and Vector Applications

<p>In the diagram, $\overrightarrow{OA}$ and $\overrightarrow{OB}$ represent vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. $X$, $Y$ and $Z$ are points such that&nbsp;$\overrightarrow{OX} = \frac{3}{2}\overrightarrow{OB}$, $\overrightarrow{AY} = \frac{2}{3}\overrightarrow{AB}$ and $\overrightarrow{OZ} = \lambda\overrightarrow{OA}$.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">Express $\overrightarrow{OX}$ and $\overrightarrow{OY}$ in terms of $\mathbf{a}$ and/or $\mathbf{b}$.</div></li><li><div class="list-item-paragraph">Express $\overrightarrow{XZ}$ in terms of $\lambda$, $\mathbf{a}$ and $\mathbf{b}$.</div></li><li><div class="list-item-paragraph">Given that&nbsp;$\lambda = \frac{3}{5}$, find the ratio of $XY : XZ$.</div></li><li><div class="list-item-paragraph">Find $\frac{\text{area of}\hspace{0.5em} \Delta OYZ}{\text{area of}\hspace{0.5em} \Delta AYZ}$.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768035372078-ucv5vvm15uk.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMzUzNzIwNzgtdWN2NXZ2bTE1dWsucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.j08avltgmJfGS_tsXhjSGTt_SA9FG1LWCx1jzk5gjAs" style="max-width: min(200px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, $\overrightarrow{OA}$ and $\overrightarrow{OB}$ represent vectors $\mathbf{a}$ and $\mathbf{b}$ respectively. $X$, $Y$ and $Z$ are points such that $\overrightarrow{OX} = \frac{3}{2}\overrightarrow{OB}$, $\overrightarrow{AY} = \frac{2}{3}\overrightarrow{AB}$ and $\overrightarrow{OZ} = \lambda\overrightarrow{OA}$.

Evaluating Vector Sums and Differences Numerically

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Given the coordinates $A(-1,3)$ and $B(0,6)$, express $\overrightarrow{AB}$ as a column vector.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence, evaluate $\left| \overrightarrow{AB} \right|$.</span></div></li></ol>

Given the coordinates $A(-1,3)$ and $B(0,6)$, express $\overrightarrow{AB}$ as a column vector.

Hence, evaluate $\left| \overrightarrow{AB} \right|$.

Magnitude of a Vector

Operations on Vectors: Addition and Subtraction

<p>In the diagram, $AQ = 3QC$, $QB = 5QX$, and $P$ is the mid-point of $AB$. $\overrightarrow{AQ} = 3\mathbf{a}$ and $\overrightarrow{AP} = \mathbf{b}$.</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/1768033586974-cxwmeravp9n.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMzM1ODY5NzQtY3h3bWVyYXZwOW4ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.XFnHGpEvf6nCg3JU0uOc3ItLSMaDzH5ZI7zrjVaRI_o" style="max-width: min(300px, 100%); height: auto; display: block;" alt="Triangle ABC with points P, Q, and intersection X" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Express, as simply as possible, in terms of $\mathbf{a}$ and or $\mathbf{b}$,</span></div><ol class="i-list"><li><div class="list-item-paragraph">$\overrightarrow{BQ}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{QX}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{CX}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{CP}$.</div></li></ol></li><li><div class="list-item-content"><span class="list-item-text">Calculate the value of</span></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">$\frac{CX}{CP}$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">$\frac{\text{area of}\hspace{0.5em} \Delta BXC}{\text{area of}\hspace{0.5em} \Delta BPX}$.</span></div></li></ol></li></ol>

In the diagram, $AQ = 3QC$, $QB = 5QX$, and $P$ is the mid-point of $AB$. $\overrightarrow{AQ} = 3\mathbf{a}$ and $\overrightarrow{AP} = \mathbf{b}$.

Express, as simply as possible, in terms of $\mathbf{a}$ and or $\mathbf{b}$,

Vector Applications: Ratios of Areas in Geometric Figures

<p>In the diagram, $WXYZ$ is a trapezium and $WX$ is parallel to $ZY$. The point $P$ on $XZ$ is such that&nbsp;$ZP:ZX=1:4$ and $4WX=3ZY$. $\overrightarrow{WX}=12\mathbf{p}$ and $\overrightarrow{WZ}=4\mathbf{q}$.</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/1768032374574-3w1yvxc5hk9.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMzIzNzQ1NzQtM3cxeXZ4YzVoazkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.3pGMHGsBxMk4xLVvG7j_tuE36D14LJqOSccxUZBsoS4" style="max-width: min(300px, 100%); height: auto; display: block;" alt="Trapezium WXYZ with vectors p and q" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Find in terms of $\mathbf{p}$ and/or $\mathbf{q}$,</span></div><ol class="i-list"><li><div class="list-item-paragraph">$\overrightarrow{ZX}$,</div></li><li><div class="list-item-paragraph">$\overrightarrow{WP}$.</div></li></ol></li><li><div class="list-item-content"><span class="list-item-text">Show that the line $XY$ is parallel to the line $WP$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the value of</span></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">$\frac{\text{area of}\hspace{0.5em} \Delta WPX}{\text{area of}\hspace{0.5em} \Delta XYZ}$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">$\frac{\text{area of}\hspace{0.5em} \Delta WZP}{\text{area of}\hspace{0.5em} \Delta XYZ}$.</span></div></li></ol></li></ol>

In the diagram, $WXYZ$ is a trapezium and $WX$ is parallel to $ZY$. The point $P$ on $XZ$ is such that $ZP:ZX=1:4$ and $4WX=3ZY$. $\overrightarrow{WX}=12\mathbf{p}$ and $\overrightarrow{WZ}=4\mathbf{q}$.

Find in terms of $\mathbf{p}$ and/or $\mathbf{q}$,

Show that the line $XY$ is parallel to the line $WP$.

Applying Vectors to Geometric Proofs and Ratio Theory

<p>Three points $O$, $P$ and $Q$ are such that $\vec{OP}=-2\mathbf{p}$ and $\vec{OQ}=3\mathbf{q}$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Express $\vec{PQ}$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Another point $R$ is such that $\vec{QR}=2\mathbf{q}-3\mathbf{p}$, find $\vec{OR}$ in terms of $\mathbf{p}$ and $\mathbf{q}$.</span></div></li></ol>

Three points $O$, $P$ and $Q$ are such that $\vec{OP}=-2\mathbf{p}$ and $\vec{OQ}=3\mathbf{q}$.

Another point $R$ is such that $\vec{QR}=2\mathbf{q}-3\mathbf{p}$, find $\vec{OR}$ in terms of $\mathbf{p}$ and $\mathbf{q}$.

Vector Addition

Calculating Directed Line Segments from Position Vectors

<p>Three points $O$, $P$ and $Q$ are such that $\vec{OP}=3\mathbf{p}$ and $\vec{OQ}=6\mathbf{q}$.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Express $\vec{PQ}$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Another point $R$ is such that $\vec{QR}=2\mathbf{q}-\mathbf{p}$, find $\vec{OR}$ in terms of $\mathbf{p}$ and $\mathbf{q}$.</span></div></li></ol>

Three points $O$, $P$ and $Q$ are such that $\vec{OP}=3\mathbf{p}$ and $\vec{OQ}=6\mathbf{q}$.

Another point $R$ is such that $\vec{QR}=2\mathbf{q}-\mathbf{p}$, find $\vec{OR}$ in terms of $\mathbf{p}$ and $\mathbf{q}$.

Introduction to Origin-Based Position Vectors

<p>Express the vector that is marked with double arrows with the given vectors in each of the diagram below.</p>
<ol class="abc-list"><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768028867500-3ld70hdbk1l.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjg4Njc1MDAtM2xkNzBoZGJrMWwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.j6O2YpkTk4fd2WEREOV57kZmiHweQEYz2tPyyFBZEY8" alt="List item image" class="alignleft custom-width" style="max-width: min(170px, 100%); height: auto;" /></div></li><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768028871324-k1df32ephq8.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjg4NzEzMjQtazFkZjMyZXBocTgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.X6CLyTLV0kdZkn1G5AY9i4yv0iH3oPYAGqn5Ydp3qaM" alt="List item image" class="alignleft custom-width" style="max-width: min(170px, 100%); height: auto;" /></div></li></ol>

Express the vector that is marked with double arrows with the given vectors in each of the diagram below.

Vector Algebra

Numeric Subtraction of 2D Column Vectors

<p>Express the vector that is marked with double arrows with the given vectors in each of the diagram below.</p>
<ol class="abc-list"><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768028557410-7x9i9vc52q.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjg1NTc0MTAtN3g5aTl2YzUycS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0OTEwLCJleHAiOjE3OTM1MDUyMTB9.d_OmrksuvixKhfDUhy5htYrywqcdyk4XtgdZZpZpGF8" alt="List item image" class="alignleft custom-width" style="max-width: min(160px, 100%); height: auto;" /></div></li><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768028563584-rxifa3cj44i.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjg1NjM1ODQtcnhpZmEzY2o0NGkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.LhP_BfTqORnQGD4HysH_cYAaTuE89Qg5h5NQ9G2p9J0" alt="List item image" class="alignleft custom-width" style="max-width: min(160px, 100%); height: auto;" /></div></li><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768028570492-jysr2dhuxij.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjg1NzA0OTItanlzcjJkaHV4aWoucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.JifJPb2Wr719NccHVBmDNBQ1U6_RjP_MrlJxTOMXJhQ" alt="List item image" class="alignleft custom-width" style="max-width: min(120px, 100%); height: auto;" /></div></li></ol>

Negative Vectors

Algebraic Methods for Vector Subtraction

<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/1768028200059-ds6p82abwbu.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjgyMDAwNTktZHM2cDgyYWJ3YnUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ._s6-hXcrsCQZEYxDvbTaSjnawcMlK3B8_ctp0h8P68s" style="max-width: min(500px, 100%); height: auto; display: block;" alt="Square grid with x and y axes" class="aligncenter custom-width" /></figure></div>
<p>Given that column vectors $\mathbf{a} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 10 \\ -4 \end{pmatrix}$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Draw, on the $1$ unit by $1$ unit square grid provided, the relationship between $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{a} - \mathbf{b}$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the column vector of $\mathbf{a} - \mathbf{b}$ numerically.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the value of $|\mathbf{a}|$, $|\mathbf{b}|$ and $|\mathbf{a} - \mathbf{b}|$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Is $|\mathbf{a} - \mathbf{b}| = |\mathbf{a}| - |\mathbf{b}|$? Explain your answer.</span></div></li></ol>

Given that column vectors $\mathbf{a} = \begin{pmatrix} 3 \\ -4 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 10 \\ -4 \end{pmatrix}$.

Draw, on the $1$ unit by $1$ unit square grid provided, the relationship between $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{a} - \mathbf{b}$.

Find the column vector of $\mathbf{a} - \mathbf{b}$ numerically.

Find the value of $|\mathbf{a}|$, $|\mathbf{b}|$ and $|\mathbf{a} - \mathbf{b}|$.

Is $|\mathbf{a} - \mathbf{b}| = |\mathbf{a}| - |\mathbf{b}|$? Explain your answer.

Numerical Analysis of Vector Subtraction in 2D

<p>Express the vector that is marked with double arrows with the given vectors in each of the diagram below.</p>
<ol class="abc-list"><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768027704154-8aqp4teh127.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjc3MDQxNTQtOGFxcDR0ZWgxMjcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.ocjuEkXKB3SKdnSWvFzuQFIP2h_V7k2sOtWRtO5G55k" alt="List item image" class="alignleft custom-width" style="max-width: min(160px, 100%); height: auto;" /></div></li><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768027717195-favg1u8hwhm.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjc3MTcxOTUtZmF2ZzF1OGh3aG0ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.EjQeg3UoOrjqlDuN759j8k9OlXLQd2TQ39QQJHK0KxE" alt="List item image" class="alignleft custom-width" style="max-width: min(160px, 100%); height: auto;" /></div></li><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768027723417-rpphzqpdl3.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjc3MjM0MTctcnBwaHpxcGRsMy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0OTEwLCJleHAiOjE3OTM1MDUyMTB9.KAdGHLfPZDLAiTWE2TLuEM2lN--Bi9WYfRBT6fl9Ys4" alt="List item image" class="alignleft custom-width" style="max-width: min(160px, 100%); height: auto;" /></div></li></ol>

Visualizing Vector Subtraction on a Cartesian Grid

<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/1768027327123-hpc54ghptfv.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgwMjczMjcxMjMtaHBjNTRnaHB0ZnYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDkxMCwiZXhwIjoxNzkzNTA1MjEwfQ.PcJ4zQML730uqbJII9y1jbhZHVZYPePe_5HlOyRg9lY" style="max-width: min(450px, 100%); height: auto; display: block;" alt="Square grid with x and y axes" class="aligncenter custom-width" /></figure></div>
<p>Given that column vectors $\mathbf{a} = \begin{pmatrix} -3 \\ 7 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Draw, on the $1$ unit by $1$ unit square grid provided, the relationship between $\mathbf{a}$, $\mathbf{b}$ and $\mathbf{a} - \mathbf{b}$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the column vector of $\mathbf{a} - \mathbf{b}$ numerically.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the value of $|\mathbf{a}|$, $|\mathbf{b}|$ and $|\mathbf{a} - \mathbf{b}|$.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Is $|\mathbf{a} - \mathbf{b}| = |\mathbf{a}| - |\mathbf{b}|$? Explain your answer.</span></div></li></ol>

Given that column vectors $\mathbf{a} = \begin{pmatrix} -3 \\ 7 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}$.

Arithmetic of Column Vectors: Subtraction and Resultants

<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 $AC:AB=\lambda :1$.

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