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

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#### (၂) တကၠသိုလ္၀င္တန္း စာေမးပြဲ ေမးခြန္းေဟာင္းမ်ား စုဆည္းမႈ

မိုးေကာင္းၿမိဳ႕   ၊ပိုင္ကိုယ္ပိုင္အထက္တန္းေက်ာင္း၊ တည္ေထာင္သူ ဆရာေမာင္ခ်မ္းသာ မွ ေစတနာေကာင္း မ်ားစြာျဖင့္ စုစည္းေပးထားေသာ တကၠသိုလ္၀င္တန္း ေက်ာင္းသား လက္ဆြဲ စာအုပ္ေကာင္း တစ္အုပ္ဟု ဆိုခ်င္ပါသည္။

ပြဲမ၀င္ခင္ အျပင္ကက်င္းပ ဆိုသည့္စကားအတိုင္း  တကၠသိုလ္၀င္တန္း စာေမးပြဲကို ၀င္ေရာက္ ေျဖဆို ၾကမည့္ ေက်ာင္းသား/သူမ်ားသည္ စားေမးပြဲ အသီးသီးတြင္ မည္သို႔ေသာ ေမးခြန္းမ်ားကို ေမးခဲ့ဘူး သည္ ဆိုသည္ကို ေသခ်ာစြာ ေလ့လာမွတ္သား ေလ့က်င့္ထားရန္ အေရးႀကီးသည္ဆိုသည္မွာ အထူးေျပာရန္ မလိုဟု ထင္မိပါသည္။

ယေန႔စာအုပ္ေပါင္းမ်ားစြာ ထြက္ရွိေနေသာ္လည္း လက္ရွိ ကိုယ္ပိုင္ေက်ာင္းကို ကိုယ္တည္တည္ေထာင္ၿပီး ဘာသာရပ္ကို ကိုယ္တိုင္သင္ၾကားေနသည့္ ဆရာတစ္ေယာက္၏ အေတြ႕အႀကံဳႏွင့္ ေစတနာေပါင္းစပ္ ထားေသာ စာအုပ္က ေက်ာင္းသားတို႔ အတြက္ မည္မွ်အေထာက္အကူ ျပဳႏိုင္သည္ဆိုသည္မွာ အထူးညႊန္းဆို ဖြယ္မရွိၿပီ။

ေမးခြန္းေဟာင္းမ်ားကို အခန္းအလိုက္၊ ႏွစ္အလိုက္ ပံုစံတူမ်ားကို စနစ္တက် က်စ္က်စ္လ်စ္လ်စ္ စုဆည္းမႈက ေက်ာင္းသားတို႔ လက္ဆြဲျပဳ အားထားရမည့္ စာအုပ္ေကာင္း တစ္အုပ္ ျဖစ္ပါေၾကာင္း...။

ဆက္သြယ္မွာ ယူရန္

ဆရာ ေမာင္ခ်မ္းသာ
ဆရာလွ
ဆရာ တင္ႏိုင္၀င္း (ေရႊမိုး Online စာေပ)
ဆရာမ ေဒၚခ်ယ္ေျမ့ခင္

(သက္ဆိုင္ရာ ဆရာမ်ား၏ နာမည္ကို ႏွိပ္လိုက္လွ်င္ ၎တို႔၏ facbook profile ကို ေရာက္သြားၿပီး Messanger မွာ မွာယူႏိုင္ပါသည္။)

#### (၃) စက္၀ိုင္း

MTG (Mathematics Teachers' Group) က စီစဥ္ျပဳစုခဲ့ေသာ တကၠသိုလ္၀င္တန္း သခၤ်ာ အခန္း (8) စာအုပ္ျဖစ္ပါသည္။ စက္၀ိုင္းဆိုသည္ႏွင့့္ ေက်ာင္းသားမ်ား အ၀ိုင္းလည္ခ်င္ ေနသည္မွာ သင္ၾကားေနေသာ ဆရာမ်ား ကိုယ္တိုင္ ႀကံဳေတြ႕ေနရသည့္ ျပသနာတရပ္ ျဖစ္ပါသည္။ အေၾကာင္းအရင္းက အေျခခံ အားနည္းျခင္း၊ ကိုယ္ေလ့က်င့္မႈ အားနည္းျခင္း၊ သင္ခန္းစာ အေပၚ ေကာင္းစြာနားမလည္ျခင္း၊ စသည့္ အခက္အခဲမ်ားေၾကာင့္ ေက်ာင္းသားတို႔ ေျဖဆိုရန္ စိုးရိမ္တတ္ေသာ သင္ခန္းစာ ျဖစ္ပါသည္။

ယင္း အခက္အခဲမ်ားကို ေျဖရွင္းေပးႏိုင္ရန္ အတြက္ ဤစာအုပ္ကို MTG မွ စီစဥ္ျပဳစု ခဲ့ပါသည္။ အလယ္တန္း အဆင့္တြင္ သင္ယူၿပီးေသာ geometry ဆိုင္ရာ သီအိုရမ္ ႏွင့္အဓိပၸာယ္ သတ္မွတ္ခ်က္မ်ား၊ မွန္ကန္ခ်က္မ်ား ၊ ဘာသာရပ္ဆိုင္ရာ အေခၚအေ၀ၚမ်ား (terminology) မ်ားကို စနစ္တက် ထည့္သြင္း ေပးထားၿပီး သင္ရိုးပါ အခန္း (၈) ေလ့က်င့္ခန္းႏွင့္ ဥပမာပုစာၦမ်ား အျပင္ ထပ္ဆင့္ေလ့လာသင့္သည့္ ေမးခြန္းေကာင္းမ်ားကိုလည္း ေပါင္းစပ္အားျဖည့္ ထားသျဖင့္ ဆရာ ေက်ာင္းသား လက္ကိုင္ထားသင့္ေသာ စာအုပ္ေကာင္း တစ္အုပ္ျဖစ္သည္ ဟု ဆိုခ်င္ပါသည္။

ဆက္သြယ္မွာ ယူရန္

ဆရာလွ

(သက္ဆိုင္ရာ ဆရာမ်ား၏ နာမည္ကို ႏွိပ္လိုက္လွ်င္ ၎တို႔၏ facbook profile ကို ေရာက္သြားၿပီး Messanger မွာ မွာယူႏိုင္ပါသည္။)
1.        Two ships leave a port at the same time. The first ship sails on a bearing of $\displaystyle 032{}^\circ$ at $\displaystyle 16$ km/h and the second on a bearing of $\displaystyle 122{}^\circ$ at $\displaystyle 24$ km/h. How far apart are they after $\displaystyle 2.5$ hours?

2.       The bearing from point $\displaystyle A$ to point $\displaystyle B$ is $\displaystyle S\ 55{}^\circ E$ and from a point $\displaystyle B$ to point $\displaystyle C$ is $\displaystyle N\ 35{}^\circ E$. If a ship sails from $\displaystyle A$ to $\displaystyle B$, a distance of $\displaystyle 81$ km, and from $\displaystyle B$ to $\displaystyle C$, a distance of $\displaystyle 74$ km, how far is it from $\displaystyle A$ to $\displaystyle C$?

3.        A pilot flew $\displaystyle 320$ miles at $\displaystyle 068{}^\circ 1{0}'$, then $\displaystyle 540$ miles at $\displaystyle 158{}^\circ 1{0}'$. The pilot then flew back to his starting point. Find the length of the last flight.

4.        A ship leaves a harbour on a bearing of $\displaystyle S\ 55{}^\circ E$ and travels for $\displaystyle 120$ km. It then turns and continues on a bearing of $\displaystyle N\ 35{}^\circ E$ for $\displaystyle 62$ km. How far is the ship from the harbour.

5.        A boy could have walked along a straight road that heads $\displaystyle N\ 30{}^\circ 3{7}' E$, but, instead, he walked due east $\displaystyle 153.9$ yards and then due north to the end of the road. How far out of his way did the boy walk?

How far out of his way = (လမ္းအတိုင္းသြားရင္ ေလ်ွာက္ရမယ့္ အကြာအေ၀းထက္) ဘယ္ေလာက္ ပိုေလွ်ာက္ခဲ့ရလဲ

6.        Two fishing boats start from a port. One travels $\displaystyle 𝟏𝟓$ nautical miles per hour on a bearing of $\displaystyle 025{}^\circ$ and the other travels $\displaystyle 𝟏𝟖$ nautical miles per hour on a bearing of $\displaystyle 𝟏𝟎𝟎°$. Assuming each maintains its course and speed, how far apart will the fishing boats be after two hours?

7.        A plane leaves from Yangon International Airport and travels due west at $\displaystyle 570$ mi/hr. Another plane leaves $\displaystyle 20$ minutes later and travels $\displaystyle 22{}^\circ$ west of north at the rate of 585 mi/h. How far apart are they 40 minutes after the second plane leaves.

8.        An airplane travels on a bearing of $\displaystyle 200{}^\circ$ for $\displaystyle 𝟏𝟓𝟎𝟎$ miles and then changes to a bearing of $\displaystyle 250{}^\circ$ and travels an additional $\displaystyle 𝟓𝟎𝟎$ miles. How far is the airplane from its starting point?

9.        Two runners start from the same point. The first one heading due north at a constant speed of $\displaystyle 12$ km/h. After $\displaystyle 2.5$ hours, the second runner is at a bearing of $\displaystyle 110{}^\circ$ from his starting point and 166.8 km away from the first.

(a) How far is the second runner from the starting point?

(b) How fast is the second runner runs on average?

10.       Yoon Yoon lives $\displaystyle 𝟐$ miles west of her school and her friend May Kyaw lives $\displaystyle 𝟑$ miles directly northeast of her.

(a) Draw a diagram representing this situation.

(b) How far does May Kyaw live from school? (c) What is the direction of school from May Kyaw's house?

11.       The tallest trees in the world are taller than a football field is long. Find the height of one of these trees, given the information in the figure.

12.       Two lighthouses are $\displaystyle 𝟑𝟎$ miles apart on each side of shorelines running north and south, as shown. Each lighthouse keeper spots a boat in the distance. One lighthouse keeper notes the location of the boat as $\displaystyle 40{}^\circ$ east of south, and the other lighthouse keeper marks the boat as $\displaystyle 32{}^\circ$ west of south. What is the distance from the boat to each of the lighthouses at the time it was spotted?

13.       Two radar stations located $\displaystyle 20$ miles apart both detect a UFO between them. The angle of elevation measured by the first station is $\displaystyle 35$ degrees. The angle of elevation measured by the second station is $\displaystyle 15$ degrees. What is the altitude of the UFO?

UFO = undefined flying object = ပန္းကန္ျပားပ်ံ

14.       To measure the height of a hill, a surveyour measures the angle of elevation to the top of the hill to be $\displaystyle 24$ degrees. He then moves back $\displaystyle 200$ feet and measures the angle of elevation to be $\displaystyle 22$ degrees. Find the height of the hill.

15.       A bridge is built across a small lake from a gazebo to a dock. The bearing from the gazebo to the dock is $\displaystyle S\ 41{}^\circ W$. From a tree $\displaystyle 100$ meters from the gazebo, the bearings to the gazebo and the dock are $\displaystyle S\ 74{}^\circ E$ and $\displaystyle S\ 28{}^\circ E$, respectively. Find the distance from the gazebo to the dock.

Gazebo = အပန္းေျဖစံအိမ္

16.       A boat is traveling due east parallel to the shoreline at a speed of $\displaystyle 10$ miles per hour. At a given time, the bearing to a lighthouse is $\displaystyle S\ 70{}^\circ E$, and $\displaystyle 15$ minutes later the bearing is $\displaystyle S\ 63{}^\circ E$. The lighthouse is located at the shoreline. What is the distance from the boat to the shoreline?

shoreline = ကမ္းပါး

17.       On a small lake, a boy swims from point of $\displaystyle A$ to point of $\displaystyle B$ at a bearing of $\displaystyle N\ 28{}^\circ E$, then to point of $\displaystyle C$ at a bearing of $\displaystyle N\ 58{}^\circ W$, and finally back to point of $\displaystyle A$, as shown in the figure below. Point of $\displaystyle C$ lies of $\displaystyle 800$ meters directly north of point of $\displaystyle A$. Find the total distance that a boy swims.