# Problem Study : Position Vectors

Problems

1.       Relative to an origin $\displaystyle O$, the position vectors of the points $\displaystyle A$ and $\displaystyle B$ are $\displaystyle 2\hat{\text{i}}-3\hat{\text{j}}$ and $\displaystyle 11\hat{\text{i}}+42\hat{\text{j}}$ respectively.

(i) Write down an expression for $\displaystyle \overrightarrow{{AB}}$.

The point C lies on AB such that such that $\displaystyle \overrightarrow{{AC}}=\frac{1}{3}\overrightarrow{{AB}}$.

(ii) Find the length of $\displaystyle \overrightarrow{{OC}}$.

The point D lies on $\displaystyle \overrightarrow{{OA}}$ such that $\displaystyle \overrightarrow{{DC}}$ is parallel to $\displaystyle \overrightarrow{{OB}}$.

(iii) Find the position vector of $\displaystyle D$.

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \ \overrightarrow{{OA}}=2\widehat{\text{i}}-3\widehat{\text{j}},\,\\\\\ \ \ \ \ \ \overrightarrow{{OB}}=11\widehat{\text{i}}+42\widehat{\text{j}}\\\\(\text{i})\ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ =\left( {11\widehat{\text{i}}+42\widehat{\text{j}}} \right)-\left( {2\widehat{\text{i}}-3\widehat{\text{j}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ =9\widehat{\text{i}}+45\widehat{\text{j}}\end{array}$

$\displaystyle (\text{ii})\ \ \ \overrightarrow{{AC}}=\frac{1}{3}\overrightarrow{{AB}}$

$\displaystyle \ \ \ \ \ \ \ \overrightarrow{{OC}}-\overrightarrow{{OA}}=\frac{1}{3}\overrightarrow{{AB}}$

$\displaystyle \ \ \ \ \ \ \ \overrightarrow{{OC}}=\frac{1}{3}\overrightarrow{{AB}}+\overrightarrow{{OA}}$$\displaystyle \therefore \ \ \ \ \ \overrightarrow{{OC}}=\frac{1}{3}\left( {9\widehat{\text{i}}+45\widehat{\text{j}}} \right)+\left( {2\widehat{\text{i}}-3\widehat{\text{j}}} \right)=5\widehat{\text{i}}+12\widehat{\text{j}}$

$\displaystyle \begin{array}{l}\therefore \ \ \ \ \ \text{Length of }\overrightarrow{{OC}}=\left| {\overrightarrow{{OC}}} \right|=\sqrt{{{{5}^{2}}+{{{12}}^{2}}}}=13\\\\(\text{iii})\ \text{D lies on }\overrightarrow{{OA}}\ \ \!\![\!\!\text{ given }\!\!]\!\!\text{ }\\\\\therefore \ \ \ \ \ \text{Let}\ \text{ }\ \overrightarrow{{OD}}=k\overrightarrow{{OA}}=2k\widehat{\text{i}}-3k\widehat{\text{j}}\\\\\ \ \ \ \ \ \ \overrightarrow{{DC}}\text{ is parallel }\overrightarrow{{OB}},\ \text{ }\!\![\!\!\text{ given }\!\!]\!\!\text{ }\\\\\therefore \ \ \ \ \ \text{Let}\ \text{ }\ \overrightarrow{{DC}}=h\overrightarrow{{OB}}\\\\\therefore \ \ \ \ \ \overrightarrow{{OC}}-\overrightarrow{{OD}}=h\overrightarrow{{OB}}\\\\\therefore \ \ \ \ \ 5\widehat{\text{i}}+12\widehat{\text{j}}-\left( {2k\widehat{\text{i}}-3k\widehat{\text{j}}} \right)=11h\widehat{\text{i}}+42h\widehat{\text{j}}\\\\\therefore \ \ \ \ \ \left( {5-2k} \right)\widehat{\text{i}}+\left( {12+3k} \right)\widehat{\text{j}}=11h\widehat{\text{i}}+42h\widehat{\text{j}}\\\\\therefore \ \ \ \ \ 5-2k=11h\Rightarrow 11h+2k=5\ \ \ ---(1)\\\\\therefore \ \ \ \ \ 12+3k=42h\Rightarrow 14h-k=4\ \ ---(2)\\\\\ \ \ \ \ \ \ \text{Solving equations (1) and (2),}\end{array}$

$\displaystyle \ \ \ \ \ \ \ h=\frac{1}{3}\ \text{and }k=\frac{2}{3}$

$\displaystyle \therefore \ \ \ \ \ \overrightarrow{{OD}}=\frac{4}{3}\widehat{\text{i}}-2\widehat{\text{j}}$

2.       The figure shows points $\displaystyle A, B$ and $\displaystyle C$ with position vectors $\displaystyle \vec{a}, \vec{b}$, and $\displaystyle \vec{c}$ respectively, relative to an origin $\displaystyle O$. The point $\displaystyle P$ lies on $\displaystyle AB$ such that $\displaystyle AP:AB=3:4$. The point $\displaystyle Q$ lies on $\displaystyle OC$ such that $\displaystyle OQ:QC=2:3$.

(i) Express $\displaystyle \overrightarrow{{AP}}$ in terms of $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ and hence show that $\displaystyle \overrightarrow{{OP}}=\frac{1}{4}(\vec{a}+3\vec{b})$.

(ii) Find $\displaystyle \overrightarrow{{PQ}}$ in terms of $\displaystyle \vec{a}, \vec{b}$, and $\displaystyle \vec{c}$.

(iii) Given that $\displaystyle 5\overrightarrow{{PQ}}=6\overrightarrow{{BC}}$, find $\displaystyle \vec{c}$ in terms of $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$.

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$\displaystyle \ \ \ \ \ \ \text{Since }AP:AB=3:4\ \text{and }A,P\operatorname{and}B\ \text{are collinear}\text{.}$

$\displaystyle \text{(i)}\ \ \ \overrightarrow{{AP}}=\frac{3}{4}\overrightarrow{{AB}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{3}{4}\left( {\overrightarrow{{OB}}-\overrightarrow{{OA}}} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{3}{4}\vec{b}-\frac{3}{4}\vec{a}$

$\displaystyle \therefore \ \ \ \ \ \overrightarrow{{OP}}=\overrightarrow{{OA}}+\overrightarrow{{AP}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\vec{a}+\frac{3}{4}\vec{b}-\frac{3}{4}\vec{a}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{4}(\vec{a}+3\vec{b})$

$\displaystyle \text{(ii)}\ \ \ Q\ \text{lies on }AC\ \text{and }OQ:QC=2:3$

$\displaystyle \therefore \ \ \ \ \ OQ:OC=2:5$

$\displaystyle \therefore \ \ \ \ \ \overrightarrow{{OQ}}=\frac{2}{5}\overrightarrow{{OC}}=\frac{2}{5}\vec{c}$

$\displaystyle \therefore \ \ \ \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\overrightarrow{{OP}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{2}{5}\vec{c}-\frac{1}{4}(\vec{a}+3\vec{b})$

$\displaystyle \text{(iii)}\ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OP}}=\vec{c}-\vec{b}$

$\displaystyle \ \ \ \ \ \ \ 5\overrightarrow{{PQ}}=6\overrightarrow{{BC}}\ \ \ \ \ \text{ }\!\![\!\!\text{ given }\!\!]\!\!\text{ }$

$\displaystyle \ \ \ \ \ \ \ 5\left( {\frac{2}{5}\vec{c}-\frac{1}{4}(\vec{a}+3\vec{b})} \right)=6\left( {\vec{c}-\vec{b}} \right)$

$\displaystyle \ \ \ \ \ \ \ 2\vec{c}-\frac{5}{4}(\vec{a}+3\vec{b})=6\vec{c}-6\vec{b}$

$\displaystyle \therefore \ \ \ \ \ 4\vec{c}=6\vec{b}-\frac{5}{4}(\vec{a}+3\vec{b})$

$\displaystyle \therefore \ \ \ \ \ 4\vec{c}=\frac{1}{4}\left( {9\vec{b}-5\vec{a}} \right)$

$\displaystyle \therefore \ \ \ \ \ \vec{c}=\frac{1}{{16}}\left( {9\vec{b}-5\vec{a}} \right)$

3.       Given that $\displaystyle \vec{p}=\left( {\begin{array}{*{20}{c}} 2 \\ {-5} \end{array}} \right)$ and $\displaystyle \vec{q}=\left( {\begin{array}{*{20}{c}} 1 \\ {-3} \end{array}} \right)$, If $\displaystyle \vec{r} =3\vec{p} - 4\vec{q}$, find the unit vector of $\displaystyle \vec{r}$.

Show/Hide Solution
$\displaystyle \ \ \ \ \ \vec{p}=\left( {\begin{array}{*{20}{c}} 2 \\ {-5} \end{array}} \right),\ \ \vec{q}=\left( {\begin{array}{*{20}{c}} 1 \\ {-3} \end{array}} \right)$

$\displaystyle \ \ \ \ \ \vec{r}=3\vec{p}-4\vec{q}\ \ \ \ \ [\text{given}]$

$\displaystyle \ \ \ \ \ \vec{r}=3\left( {\begin{array}{*{20}{c}} 2 \\ {-5} \end{array}} \right)-4\left( {\begin{array}{*{20}{c}} 1 \\ {-3} \end{array}} \right)$

$\displaystyle \ \ \ \ \ \vec{r}=\left( {\begin{array}{*{20}{c}} 6 \\ {-15} \end{array}} \right)+\left( {\begin{array}{*{20}{c}} {-4} \\ {12} \end{array}} \right)$

$\displaystyle \ \ \ \ \ \vec{r}=\left( {\begin{array}{*{20}{c}} 2 \\ {-3} \end{array}} \right)$

$\displaystyle \therefore \ \ \ \left| {\vec{r}} \right|=\sqrt{{{{2}^{2}}+{{{(-3)}}^{2}}}}=\sqrt{{13}}$

$\displaystyle \therefore \ \ \ \text{unit vector of }\vec{r}=\frac{{\vec{r}}}{{\left| {\vec{r}} \right|}}=\frac{1}{{\sqrt{{13}}}}\left( {\begin{array}{*{20}{c}} 2 \\ {-3} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {\frac{2}{{\sqrt{{13}}}}} \\ {\frac{{-3}}{{\sqrt{{13}}}}} \end{array}} \right)$

4.       Vectors $\displaystyle \hat{\text{i}}$ and $\displaystyle \hat{\text{j}}$ are unit vectors parallel to the $\displaystyle x$-axis and $\displaystyle y$-axis respectively.

The vector $\displaystyle \vec{p}$ has a magnitude of 39 units and has the same direction as $\displaystyle -10\hat{\text{i}}+24\hat{\text{j}}$.

(i) Find $\displaystyle \vec{p}$ in terms of $\displaystyle \hat{\text{i}}$ and $\displaystyle \hat{\text{j}}$ .

(ii) Find the vector $\displaystyle \vec{q}$ such that $\displaystyle 2\vec{p}+\vec{q}$ is parallel to the positive $\displaystyle y$-axis and has a magnitude of 12 units.

(iii) Express $\displaystyle \left| {\vec{q}} \right|=k\sqrt{5}$, where $\displaystyle k$ is an integer and hence find the value of $\displaystyle k$ .

Show/Hide Solution
$\displaystyle \ \ \ \ \ \ \ \text{Let}\ \vec{r}=-10\widehat{\text{i}}+24\widehat{\text{j}}$

$\displaystyle \therefore \ \ \ \ \ \ \left| {\vec{r}} \right|=\sqrt{{{{{(-10)}}^{2}}+{{{24}}^{2}}}}=26$

$\displaystyle \therefore \ \ \ \ \ \ \hat{r}=\frac{{\vec{r}}}{{\left| {\vec{r}} \right|}}=\frac{1}{{26}}\left( {-10\widehat{\text{i}}+24\widehat{\text{j}}} \right)$

$\displaystyle (\text{i})\ \ \ \ \vec{p}=39\hat{r}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ =\frac{3}{2}\left( {-10\widehat{\text{i}}+24\widehat{\text{j}}} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ =-15\widehat{\text{i}}+36\widehat{\text{j}}$

$\displaystyle (\text{ii})\$ Since$\displaystyle 2\vec{p}+\vec{q}$ is parallel to the positive $\displaystyle y$-axis and has a magnitude of $\displaystyle 12$ units.

$\displaystyle \ \ \ \ \ \ \ \ 2\vec{p}+\vec{q}=12\widehat{\text{j}}$

$\displaystyle \therefore \ \ \ \ \ \ 2\left( {-15\widehat{\text{i}}+36\widehat{\text{j}}} \right)+\vec{q}=12\widehat{\text{j}}$

$\displaystyle \therefore \ \ \ \ \ \ \vec{q}=30\widehat{\text{i}}-60\widehat{\text{j}}$

$\displaystyle \therefore \ \ \ \ \ \ \left| {\vec{r}} \right|=\sqrt{{{{{30}}^{2}}+{{{(-60)}}^{2}}}}=30\sqrt{5}$

$\displaystyle \ \ \ \ \ \ \ \ \left| {\vec{r}} \right|=k\sqrt{5}\ \ \ \ [\text{given}]$

$\displaystyle \therefore \ \ \ \ \ \ k=30$

5.       The diagram shows a figure $\displaystyle OABC$, where $\displaystyle \overrightarrow{{OA}}=\vec{a}$, $\displaystyle \overrightarrow{{OB}}=\vec{b}$ and $\displaystyle \overrightarrow{{OC}}=\vec{c}$. The lines $\displaystyle AC$ and $\displaystyle OB$ intersect at the point $\displaystyle M$ where $\displaystyle M$ is the midpoint of the line $\displaystyle AC$.

$\displaystyle \text{(i)}$ Find, in terms of $\displaystyle \vec{a}$ and $\displaystyle \vec{c}$, the vector $\displaystyle \overrightarrow{{OM}}$.

$\displaystyle \text{(ii)}$ Given that $\displaystyle OM : MB = 2 : 3$, find $\displaystyle \vec{b}$ in terms of $\displaystyle \vec{a}$ and $\displaystyle \vec{c}$.

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \overrightarrow{{OA}}=\vec{a},\ \overrightarrow{{OB}}=\vec{b},\ \overrightarrow{{OC}}=\vec{c}\\\\\ \ \ \ \ \ \ \ \text{Since}\ M\ \text{is the midpoint of }AC,\end{array}$

$\displaystyle \text{(i) }\ \ \ \ \overrightarrow{{OM}}=\frac{1}{2}\left( {\overrightarrow{{OA}}+\overrightarrow{{OC}}} \right)$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\left( {\vec{a}+\vec{c}} \right)$

$\displaystyle \text{(ii) }\ \ \text{Since}\ \frac{{OM}}{{MB}}=\frac{2}{3},$

$\displaystyle \ \ \ \ \ \ \ \frac{{OM}}{{OB}}=\frac{2}{5}\Rightarrow \ \frac{{OB}}{{OM}}=\frac{5}{2}$

$\displaystyle \therefore \ \ \ \ \ OB=\frac{5}{2}OM$

$\displaystyle \ \ \ \ \ \ \ \text{Since}\ O,\ M,\ B\ \text{are collinear,}$

$\displaystyle \ \ \ \ \ \ \ \overrightarrow{{OB}}=\frac{5}{2}\overrightarrow{{OM}}$

$\displaystyle \therefore \ \ \ \ \ \vec{b}=\frac{5}{4}\left( {\vec{a}+\vec{c}} \right)$