Position Vectors : Practice Problems

1.        Find the unit vector in the direction of $ \displaystyle \overrightarrow{{PQ}}$ where $ \displaystyle P$ and $ \displaystyle Q$ are points $ \displaystyle (2, 3)$ and $ \displaystyle (7, – 9)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ P\operatorname{and}\ Q\ \text{are points}\ (2,3)\ \operatorname{and}\ (7,-9).\\\\\therefore \ \ \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} 2 \\ 3 \end{array}} \right)\ \operatorname{and}\ \ \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} 7 \\ {-9} \end{array}} \right)\\\\\ \ \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\ \overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 7 \\ {-9} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 2 \\ 3 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 5 \\ {-12} \end{array}} \right)\\\\\therefore \ \ \left| {\ \overrightarrow{{PQ}}} \right|=\sqrt{{{{5}^{2}}+{{{\left( {-12} \right)}}^{2}}}}=13\\\\\therefore \ \ \text{The unit vector in }\\\ \ \ \ \text{the direction of}\ \ \ \ =\displaystyle \frac{{\overrightarrow{{PQ}}}}{{\left| {\ \overrightarrow{{PQ}}} \right|}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{13}}\left( {\begin{array}{*{20}{c}} 5 \\ {-12} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {\displaystyle \frac{5}{{13}}} \\ {-\displaystyle \frac{{12}}{{13}}} \end{array}} \right)\end{array}$

2.        Given that $ \displaystyle \overrightarrow{{OP}}=\widehat{\text{i}}+2\widehat{\text{j}}$ and $ \displaystyle \overrightarrow{{OQ}}=7\widehat{\text{i}}-4\widehat{\text{j}}$. Find the position vector of a point $ \displaystyle R$ which lies on the line $ \displaystyle PQ$ such that $ \displaystyle PR : RQ = 2 : 1$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OP}}=\widehat{\text{i}}+2\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OQ}}=7\widehat{\text{i}}-4\widehat{\text{j}}\\\\\ \ \ \ \ PR:RQ=2:1\\\\\ \ \ \ \ \text{By section formula},\ \\\\\ \ \ \ \ \ \overrightarrow{{OR}}=\displaystyle \frac{{\left( {1\times \overrightarrow{{OP}}} \right)+\left( {2\times \overrightarrow{{OQ}}} \right)}}{{1+2}}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{3}\ \left[ {\widehat{\text{i}}+2\widehat{\text{j}}+2\left( {7\widehat{\text{i}}-4\widehat{\text{j}}} \right)} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ =5\widehat{\text{i}}-2\widehat{\text{j}}\end{array}$

3.        If $ \displaystyle \overrightarrow{{OA}}=-5\widehat{\text{i}}+6\widehat{\text{j}}$, $ \displaystyle \overrightarrow{{OB}}=2\widehat{\text{i}}+5\widehat{\text{j}}$ and $ \displaystyle \overrightarrow{{OC}}=9\widehat{\text{i}}+4\widehat{\text{j}}$, show that $ \displaystyle A, B$ and $ \displaystyle C$ are collinear.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=-5\widehat{\text{i}}+6\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OB}}=2\widehat{\text{i}}+5\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=9\widehat{\text{i}}+4\widehat{\text{j}}\\\ \ \ \ \ \\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {2\widehat{\text{i}}+5\widehat{\text{j}}} \right)-\left( {-5\widehat{\text{i}}+6\widehat{\text{j}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =7\widehat{\text{i}}-\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {9\widehat{\text{i}}+4\widehat{\text{j}}} \right)-\left( {2\widehat{\text{i}}+5\widehat{\text{j}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ =7\widehat{\text{i}}-\widehat{\text{j}}\\\\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{BC}}\\\\\therefore \ \ \ A,B\ \operatorname{and}\ C\ \text{are collinear}.\end{array}$

4.        Given that $ \displaystyle \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} k \\ 5 \end{array}} \right)$, $ \displaystyle \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right)$ and $ \displaystyle \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 3 \\ {11} \end{array}} \right)$. If $ \displaystyle P, Q$ and $ \displaystyle R$ are collinear, find the value of $ \displaystyle k$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} k \\ 5 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 3 \\ {11} \end{array}} \right),\\\\\therefore \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} k \\ 5 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2-k} \\ 3 \end{array}} \right)\\\\\ \ \ \ \overrightarrow{{QR}}=\overrightarrow{{OR}}-\overrightarrow{{OQ}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 3 \\ {11} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-2} \\ 8 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right)\\\\\ \ \ \ \text{By the problem,}\ \\\\\ \ \ P,\ Q\ \operatorname{and}\ R\ \text{are collinear}.\\\\\therefore \ \ \text{Let}\ \overrightarrow{{PQ}}=h\overrightarrow{{QR}}\\\\\therefore \ \ \left( {\begin{array}{*{20}{c}} {-2-k} \\ 3 \end{array}} \right)=h\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right)\\\\\ \ \ \ \left( {\begin{array}{*{20}{c}} {-2-k} \\ 3 \end{array}} \right)=\left( {\begin{array}{*{20}{c}} {5h} \\ {3h} \end{array}} \right)\\\\\therefore \ \ 3h=3\\\\\ \ \ \ h=1\\\\\ \ \ \ -2-k=5h\\\\\ \ \ k=-2-5h\\\\\ \ \ k=-7\\\ \ \ \end{array}$

5.        Using a vector method, show that the points $ \displaystyle A (– 8, 10), B (– 1, 9)$ and $ \displaystyle C (6, 8)$ are collinear and hence find the ratio $ \displaystyle AB : BC$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=\left( {\begin{array}{*{20}{c}} {-8} \\ {10} \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OB}}=\left( {\begin{array}{*{20}{c}} {-1} \\ 9 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OC}}=\left( {\begin{array}{*{20}{c}} 6 \\ 8 \end{array}} \right),\\\\\therefore \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-1} \\ 9 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-8} \\ {10} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 7 \\ {-1} \end{array}} \right)\\\\\ \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 6 \\ 8 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-1} \\ 9 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 7 \\ {-1} \end{array}} \right)\\\\\therefore \ \ \overrightarrow{{AB}}=\overrightarrow{{BC}}\\\\\ \ \ A,\ B\ \operatorname{and}\ C\ \text{are collinear and }\\\\\ \ \ AB:BC=1:1\ \end{array}$

6.        If $ \displaystyle \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right)$, $ \displaystyle \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right)$ and $ \displaystyle \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right)$, show that $ \displaystyle \Delta PQR$ is a right triangle.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OP}}=\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OQ}}=\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right),\\\\\ \ \ \ \overrightarrow{{OR}}=\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right),\\\\\therefore \ \ \overrightarrow{{PQ}}=\overrightarrow{{OQ}}-\overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2} \\ 6 \end{array}} \right)\\\\\therefore \ PQ=\sqrt{{{{{\left( {-2} \right)}}^{2}}+{{6}^{2}}}}=\sqrt{{40}}\\\\\ \ \ \ \overrightarrow{{QR}}=\overrightarrow{{OR}}-\overrightarrow{{OQ}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-5} \\ {14} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {14} \\ {-2} \end{array}} \right)\\\\\therefore \ QR=\sqrt{{{{{14}}^{2}}+{{{\left( {-2} \right)}}^{2}}}}=\sqrt{{200}}\\\\\ \overrightarrow{{PR}}=\overrightarrow{{OR}}-\overrightarrow{{OP}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 9 \\ {12} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} {-3} \\ 8 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {12} \\ 4 \end{array}} \right)\\\\\therefore \ \ PR=\sqrt{{{{{12}}^{2}}+{{4}^{2}}}}=\sqrt{{160}}\\\\\ \ \ \ P{{Q}^{2}}+P{{R}^{2}}=40+160=200\\\\\ \ \ \ Q{{R}^{2}}=200\\\\\therefore \ \ P{{Q}^{2}}+P{{R}^{2}}=\ Q{{R}^{2}}\\\\\therefore \ \ \Delta PQR\ \text{is a right triangle}\text{.}\end{array}$

7.        If the position vectors of the points $ \displaystyle A, B$ and $ \displaystyle C$ are $ \displaystyle 9\widehat{\text{i}}+6\widehat{\text{j}}$, $ \displaystyle 4\widehat{\text{i}}+3\widehat{\text{j}}$ and $\displaystyle -5\widehat{\text{i}}+8\widehat{\text{j}}$ respectively, show that the $ \displaystyle \Delta ABC$ is an obtuse triangle.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=9\widehat{\text{i}}+6\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OB}}=4\widehat{\text{i}}+3\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=-5\widehat{\text{i}}+8\widehat{\text{j}}\\\\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =-5\widehat{\text{i}}-3\widehat{\text{j}}\\\\\therefore \ \ \ A{{B}^{2}}={{\left( {-5} \right)}^{2}}+{{\left( {-3} \right)}^{2}}=34\\\\\therefore \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =-9\widehat{\text{i}}+5\widehat{\text{j}}\\\\\therefore \ \ \ B{{C}^{2}}={{\left( {-9} \right)}^{2}}+{{5}^{2}}=106\\\\\therefore \ \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =-14\widehat{\text{i}}+2\widehat{\text{j}}\\\\\therefore \ \ \ A{{C}^{2}}={{\left( {-14} \right)}^{2}}+{{2}^{2}}=200\\\\\therefore \ \ \ A{{B}^{2}}+B{{C}^{2}}=140\\\\\therefore \ \ \ A{{C}^{2}}>A{{B}^{2}}+B{{C}^{2}}\\\\\therefore \ \ \ \Delta ABC\ \text{is an obtuse triangle}\text{.}\end{array}$

8.        The position vectors of the points A, B, and C are $ \displaystyle \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right)$, $ \displaystyle \left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right)$ and $ \displaystyle \left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)$ respectively. Prove that $ \displaystyle \Delta ABC$ is an isosceles right triangle.

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$ \displaystyle \begin{array}{l}\ \ \ \overrightarrow{{OA}}=\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right),\\\\\ \ \ \overrightarrow{{OB}}=\left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right),\\\\\ \ \ \overrightarrow{{OC}}=\left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)\\\\\ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-2} \\ {-6} \end{array}} \right)\\\\\therefore \ \ AB=\sqrt{{{{{\left( {-2} \right)}}^{2}}+{{{\left( {-6} \right)}}^{2}}}}=\sqrt{{40}}\\\\\ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 3 \\ {-4} \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-4} \\ 8 \end{array}} \right)\\\\\therefore \ \ BC=\sqrt{{{{{\left( {-4} \right)}}^{2}}+{{8}^{2}}}}=\sqrt{{80}}\\\\\ \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-1} \\ 4 \end{array}} \right)-\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right)\\\\\ \ \ \ \ \ \ \ \ =\left( {\begin{array}{*{20}{c}} {-6} \\ 2 \end{array}} \right)\\\\\therefore \ \ AC=\sqrt{{{{{\left( {-6} \right)}}^{2}}+{{2}^{2}}}}=\sqrt{{40}}\\\\\therefore \ \ AB=AC\\\\\ \ \ \ A{{B}^{2}}+A{{C}^{2}}=40+40=80=B{{C}^{2}}\\\\\therefore \ \ \ \Delta ABC\ \text{is an isosceles right triangle}\text{.}\end{array}$

9.        $ \displaystyle OABC$ is a parallelogram such that $ \displaystyle \overrightarrow{{OA}}=5\widehat{\text{i}}+3\widehat{\text{j}}$ and $ \displaystyle \overrightarrow{{OC}}=-2\widehat{\text{i}}+\widehat{\text{j}}$. Find the unit vector in the direction of $ \displaystyle \ \overrightarrow{{OB}}$ and $ \displaystyle \ \overrightarrow{{AC}}$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=5\widehat{\text{i}}+3\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=-2\widehat{\text{i}}+\widehat{\text{j}}\\\\\ \ \ \ \ OABC\ \text{is a parallelogram}.\\\\\therefore \ \ \ \overrightarrow{{OB}}=\overrightarrow{{OA}}+\overrightarrow{{OC}}\ \ \ \left( {\because \text{parallelogram rule}\text{.}} \right)\\\\\ \ \ \ \ \overrightarrow{{OB}}=5\widehat{\text{i}}+3\widehat{\text{j}}-2\widehat{\text{i}}+\widehat{\text{j}}=3\widehat{\text{i}}+4\widehat{\text{j}}\\\\\therefore \ \ \ OB=\sqrt{{{{3}^{2}}+{{4}^{2}}}}=5\\\\\therefore \ \ \ \text{the unit vector in }\\\ \ \ \ \ \text{the direction of}\ \overrightarrow{{OB}}=\displaystyle \frac{{\overrightarrow{{OB}}}}{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{5}\left( {3\widehat{\text{i}}+4\widehat{\text{j}}} \right)\\\\\ \ \ \ \ \ \text{Again}\ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\ \ \\\ \ \ \ \ \overrightarrow{{AC}}=\left( {-2\widehat{\text{i}}+\widehat{\text{j}}} \right)-\left( {5\widehat{\text{i}}+3\widehat{\text{j}}} \right)=-7\widehat{\text{i}}-2\widehat{\text{j}}\\\\\therefore \ \ \ AC=\sqrt{{{{{\left( {-7} \right)}}^{2}}+{{{\left( {-2} \right)}}^{2}}}}=\sqrt{{53}}\\\\\therefore \ \ \ \text{the unit vector in }\\\ \ \ \ \ \text{the direction of}\ \overrightarrow{{AC}}=\displaystyle \frac{{\overrightarrow{{AC}}}}{{AC}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{1}{{\sqrt{{53}}}}\left( {-7\widehat{\text{i}}-2\widehat{\text{j}}} \right)\end{array}$

10.       Given that the position vectors of the points $ \displaystyle A, B$ and $ \displaystyle C$ relative to origin $ \displaystyle O$ are $ \displaystyle -\widehat{\text{i}}+\widehat{\text{j}}$, $ \displaystyle 5\widehat{\text{i}}+\widehat{\text{j}}$ and $ \displaystyle p\widehat{\text{i}}+q\widehat{\text{j}}$ respectively. If $ \displaystyle \Delta ABC$ is equilateral, find the possible values of $ \displaystyle p$ and $ \displaystyle q$.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \overrightarrow{{OA}}=-\widehat{\text{i}}+\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OB}}=5\widehat{\text{i}}+\widehat{\text{j}}\\\\\ \ \ \ \ \overrightarrow{{OC}}=p\widehat{\text{i}}+q\widehat{\text{j}}\\\\\therefore \ \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =6\widehat{\text{i}}\\\\\therefore \ \ \ AB=6\\\\\ \ \ \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =\left( {p-5} \right)\widehat{\text{i}}+\left( {q-1} \right)\widehat{\text{j}}\\\\\ \ \ \ \ BC=\sqrt{{{{{\left( {p-5} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}\\\\\ \ \ \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ \ =\left( {p+1} \right)\widehat{\text{i}}+\left( {q-1} \right)\widehat{\text{j}}\\\\\ \ \ \ \ AC=\sqrt{{{{{\left( {p+1} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}\\\\\ \ \ \ \ \text{Since }\Delta ABC\ \text{is equilateral,}\\\\\ \ \ \ \ AB=BC=AC=6.\\\\\therefore \ \ \sqrt{{{{{\left( {p-5} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}=6\\\\\ \ \ \ {{\left( {p-5} \right)}^{2}}+{{\left( {q-1} \right)}^{2}}=36\\\\\ \ \ \ {{p}^{2}}+{{q}^{2}}-10p-2q=10\ ---(1)\\\\\ \ \ \ \text{Similarly,}\\\\\ \ \ \ \sqrt{{{{{\left( {p+1} \right)}}^{2}}+{{{\left( {q-1} \right)}}^{2}}}}=6\\\\\ \ \ \ {{\left( {p+1} \right)}^{2}}+{{\left( {q-1} \right)}^{2}}=36\\\\\ \ \ \ {{p}^{2}}+{{q}^{2}}+2p-2q=34\ ---(2)\\\\\ \ \ \ \text{By (2)}-\text{(1),}\\\text{ }\\\ \ \ \ 12p=24\\\\\therefore \ \ p=2\\\\\ \ \ \ \text{Substituting}\ p=2\text{ in (1),}\\\text{ }\\\therefore \ \ 4+{{q}^{2}}-20-2q=10\\\ \\\ \ \ \ {{q}^{2}}-2q=26\\\\\ \ \ \ {{q}^{2}}-2q+1=27\\\\\ \ \ \ {{(q-1)}^{2}}=27\\\\\ \ \ \ q-1=\pm \sqrt{{27}}\\\\\ \ \ \ q=1\pm 3\sqrt{3}\end{array}$

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$.

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$ \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}$.

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$ \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$ .

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$ \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}$.

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$ \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)$

Problem Study : Geometric Vector

The diagram shows triangle $ \displaystyle OAB$ in which $ \displaystyle \overrightarrow{{OA}}=\vec{a}$ and $ \displaystyle \overrightarrow{{OB}}=\vec{b}$. The point $ \displaystyle C$ is the midpoint of $ \displaystyle OA$ and the point $ \displaystyle D$ is the midpoint of $ \displaystyle DC$.

(a) Express $ \overrightarrow{{OD}}$ in terms of $ \vec{a}$ and $ \vec{b}$.

(b) If the point E lies on AB, then show that $ \overrightarrow{{OE}}$ can be written in the form $ \displaystyle \vec{a}+k\left( {\vec{b}-\vec{a}} \right)$, where k is a constant.

      Given also that $ \displaystyle OB$ produced meets $ \displaystyle AB$ at $ \displaystyle E$,

(c) find $ \overrightarrow{{OE}}$,

(d) $ \displaystyle AB:EB=2:1$ 

Solution

$ \displaystyle \text{Since}$ $ \displaystyle D$ $ \displaystyle \text{is the midpoint of}$ $ \displaystyle BC$, $ \displaystyle \text{by midpoint formula,}$

$ \displaystyle \begin{array}{l}\overrightarrow{{OD}}=\frac{1}{2}\left( {\overrightarrow{{OC}}+\overrightarrow{{OB}}} \right)\\\\\ \ \ \ \ \ =\frac{1}{2}\left( {\frac{1}{2}\vec{a}+\vec{b}} \right)\\\\\ \ \ \ \ \ =\frac{1}{4}\vec{a}+\frac{1}{2}\vec{b}\end{array}$

$ \displaystyle \text{Let}$ $ \displaystyle E$ $ \displaystyle \text{be a point on}$ $ \displaystyle AB$ $ \displaystyle \text{such that}$ $ \displaystyle \overrightarrow{{AE}}=k\overrightarrow{{AB}}.$ 

$ \displaystyle \begin{array}{l}\therefore \overrightarrow{{OE}}=\overrightarrow{{OA}}+\overrightarrow{{AE}}\\\\\ \ \ \ \ \ \ \ =\overrightarrow{{OA}}+k\overrightarrow{{AB}}\\\\\ \ \ \ \ \ \ \ =\overrightarrow{{OA}}+k\left( {\overrightarrow{{OB}}-\overrightarrow{{OA}}} \right)\\\\\ \ \ \ \ \ \ \ =\vec{a}+k\left( {\vec{b}-\vec{a}} \right)\end{array}$

$ \displaystyle \text{Since}$ $ \displaystyle OD$ $ \displaystyle \text{produced meets}$ $ \displaystyle AB$ $ \displaystyle \text{at}$ $ \displaystyle E,$ $ \displaystyle \text{let}$ $ \displaystyle \overrightarrow{{OE}}=h\overrightarrow{{OD}}.$ 

$ \displaystyle \therefore \left( {1-k} \right)\vec{a}+k\vec{b}=\frac{h}{4}\vec{a}+\frac{h}{2}\vec{b}$

$ \displaystyle \text{Since}$ $ \displaystyle \vec{a}$ $ \displaystyle \text{and}$ $ \displaystyle \vec{b}$ $ \displaystyle \text{are not parallel and}$ $ \displaystyle \vec{a}\ne \vec{0},\vec{b}\ne\vec{0},$ 

$ \displaystyle \begin{array}{l}\ \ \ k=\frac{h}{2}\Rightarrow h=2k\\\\\ \ \ 1-k=\frac{h}{4}\Rightarrow h=4-4k\\\\\therefore 2k=4-4k\Rightarrow k=\frac{2}{3}\end{array}$ 

$ \displaystyle \text{Since }\overrightarrow{{AE}}=k\overrightarrow{{AB}},$

$ \displaystyle \begin{array}{l}\overrightarrow{{AE}}=\frac{2}{3}\overrightarrow{{AB}}\Rightarrow AE=\frac{2}{3}AB\\\\\therefore \overrightarrow{{EB}}=\frac{1}{3}\overrightarrow{{AB}}\Rightarrow EB=\frac{1}{3}AB\\\\\therefore AE:EB=\frac{2}{3}AB:\frac{1}{3}AB\\\\\therefore AE:EB=2:1\end{array}$