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) တက္ကသိုလ်ဝင်တန်း စာမေးပွဲ မေးခွန်းဟောင်းများ စုဆည်းမှု

မိုးကောင်းမြို့   ၊ပိုင်ကိုယ်ပိုင်အထက်တန်းကျောင်း၊ တည်ထောင်သူ ဆရာမောင်ချမ်းသာ မှ စေတနာကောင်း များစွာဖြင့် စုစည်းပေးထားသော တက္ကသိုလ်ဝင်တန်း ကျောင်းသား လက်ဆွဲ စာအုပ်ကောင်း တစ်အုပ်ဟု ဆိုချင်ပါသည်။

ပွဲမဝင်ခင် အပြင်ကကျင်းပ ဆိုသည့်စကားအတိုင်း  တက္ကသိုလ်ဝင်တန်း စာမေးပွဲကို ဝင်ရောက် ဖြေဆို ကြမည့် ကျောင်းသား/သူများသည် စားမေးပွဲ အသီးသီးတွင် မည်သို့သော မေးခွန်းများကို မေးခဲ့ဘူး သည် ဆိုသည်ကို သေချာစွာ လေ့လာမှတ်သား လေ့ကျင့်ထားရန် အရေးကြီးသည်ဆိုသည်မှာ အထူးပြောရန် မလိုဟု ထင်မိပါသည်။

ယနေ့စာအုပ်ပေါင်းများစွာ ထွက်ရှိနေသော်လည်း လက်ရှိ ကိုယ်ပိုင်ကျောင်းကို ကိုယ်တည်တည်ထောင်ပြီး ဘာသာရပ်ကို ကိုယ်တိုင်သင်ကြားနေသည့် ဆရာတစ်ယောက်၏ အတွေ့အကြုံနှင့် စေတနာပေါင်းစပ် ထားသော စာအုပ်က ကျောင်းသားတို့ အတွက် မည်မျှအထောက်အကူ ပြုနိုင်သည်ဆိုသည်မှာ အထူးညွှန်းဆို ဖွယ်မရှိပြီ။

မေးခွန်းဟောင်းများကို အခန်းအလိုက်၊ နှစ်အလိုက် ပုံစံတူများကို စနစ်တကျ ကျစ်ကျစ်လျစ်လျစ် စုဆည်းမှုက ကျောင်းသားတို့ လက်ဆွဲပြု အားထားရမည့် စာအုပ်ကောင်း တစ်အုပ် ဖြစ်ပါကြောင်း...။

ဆက်သွယ်မှာ ယူရန်

ဆရာ မောင်ချမ်းသာ
ဆရာလှ
ဆရာ တင်နိုင်ဝင်း (ရွှေမိုး Online စာပေ)
ဆရာမ ဒေါ်ချယ်မြေ့ခင်

(သက်ဆိုင်ရာ ဆရာများ၏ နာမည်ကို နှိပ်လိုက်လျှင် ၎င်းတို့၏ facbook profile ကို ရောက်သွားပြီး Messanger မှာ မှာယူနိုင်ပါသည်။)

#### (၃) စက်ဝိုင်း

MTG (Mathematics Teachers' Group) က စီစဉ်ပြုစုခဲ့သော တက္ကသိုလ်ဝင်တန်း သင်္ချာ အခန်း (8) စာအုပ်ဖြစ်ပါသည်။ စက်ဝိုင်းဆိုသည်နှင့် ကျောင်းသားများ အဝိုင်းလည်ချင် နေသည်မှာ သင်ကြားနေသော ဆရာများ ကိုယ်တိုင် ကြုံတွေ့နေရသည့် ပြသနာတရပ် ဖြစ်ပါသည်။ အကြောင်းအရင်းက အခြေခံ အားနည်းခြင်း၊ ကိုယ်လေ့ကျင့်မှု အားနည်းခြင်း၊ သင်ခန်းစာ အပေါ် ကောင်းစွာနားမလည်ခြင်း၊ စသည့် အခက်အခဲများကြောင့် ကျောင်းသားတို့ ဖြေဆိုရန် စိုးရိမ်တတ်သော သင်ခန်းစာ ဖြစ်ပါသည်။

ယင်း အခက်အခဲများကို ဖြေရှင်းပေးနိုင်ရန် အတွက် ဤစာအုပ်ကို MTG မှ စီစဉ်ပြုစု ခဲ့ပါသည်။ အလယ်တန်း အဆင့်တွင် သင်ယူပြီးသော geometry ဆိုင်ရာ သီအိုရမ် နှင့်အဓိပ္ပာယ် သတ်မှတ်ချက်များ၊ မှန်ကန်ချက်များ ၊ ဘာသာရပ်ဆိုင်ရာ အခေါ်အဝေါ်များ (terminology) များကို စနစ်တကျ ထည့်သွင်း ပေးထားပြီး သင်ရိုးပါ အခန်း (၈) လေ့ကျင့်ခန်းနှင့် ဥပမာပုစ္ဆာများ အပြင် ထပ်ဆင့်လေ့လာသင့်သည့် မေးခွန်းကောင်းများကိုလည်း ပေါင်းစပ်အားဖြည့် ထားသဖြင့် ဆရာ ကျောင်းသား လက်ကိုင်ထားသင့်သော စာအုပ်ကောင်း တစ်အုပ်ဖြစ်သည် ဟု ဆိုချင်ပါသည်။

ဆက်သွယ်မှာ ယူရန်

ဆရာလှ

(သက်ဆိုင်ရာ ဆရာများ၏ နာမည်ကို နှိပ်လိုက်လျှင် ၎င်းတို့၏ facbook profile ကို ရောက်သွားပြီး Messanger မှာ မှာယူနိုင်ပါသည်။)
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 = (လမ်းအတိုင်းသွားရင် လျှောက်ရမယ့် အကွာအဝေးထက်) ဘယ်လောက် ပိုလျှောက်ခဲ့ရလဲ

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 = ပန်းကန်ပြားပျံ
radar stations = ရေဒါစခန်းများ

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 = အပန်းဖြေစံအိမ်

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 = ကမ်းပါး

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.