\displaystyle \textbf{Question 1: }~\text{If }P,Q\text{ and }R\text{ are three collinear points such that }\overrightarrow{PQ}=\overrightarrow{a}\text{ and }\overrightarrow{QR}=\overrightarrow{b},\text{ find the vector }\overrightarrow{PR}.
\displaystyle \text{Answer:}
\displaystyle \text{Given: } P,Q \text{ and } R \text{ are collinear such that } \overrightarrow{PQ}=\overrightarrow{a} \text{ and } \overrightarrow{QR}=\overrightarrow{b}. \\ \text{Then } \overrightarrow{PQ}+\overrightarrow{QR}=\overrightarrow{PR}
\displaystyle \Rightarrow \overrightarrow{PR}=\overrightarrow{a}+\overrightarrow{b}

\displaystyle \textbf{Question 2: }~\text{Give a condition that three vectors }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ form the three} \\ \text{sides of a triangle. What are the other possibilities?}
\displaystyle \text{Answer:}
\displaystyle \text{Let } ABC \text{ be a triangle such that } \overrightarrow{BC}=\overrightarrow{a}, \ \overrightarrow{AB}=\overrightarrow{c} \text{ and } \overrightarrow{CA}=\overrightarrow{b}. \\ \text{Then } \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}
\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{BA}+\overrightarrow{AB}
\displaystyle [\because \ \overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{BA}]
\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{BB} \ \ [\text{Using triangle law}]
\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0} \ \ [\text{By definition of null vector}]
\displaystyle \text{Other possibilities are}
\displaystyle \text{(i) } \overrightarrow{c}+\overrightarrow{a}=\overrightarrow{b}
\displaystyle \text{(ii) } \overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}
\displaystyle \text{(iii) } \overrightarrow{b}+\overrightarrow{c}=\overrightarrow{a}

\displaystyle \textbf{Question 3: }~\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are two non-collinear vectors having the same initial point. } \\ \text{What are the vectors represented by }\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{a}-\overrightarrow{b}?
\displaystyle \text{Answer:}


\displaystyle \text{Given: } \overrightarrow{a},\overrightarrow{b} \text{ are two non-collinear vectors having the same initial point. Complete} \\ \text{the parallelogram } ABCD \text{ such that } \overrightarrow{AB}=\overrightarrow{a} \text{ and } \overrightarrow{BC}=\overrightarrow{b}
\displaystyle \text{In } \triangle ABC
\displaystyle \overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}
\displaystyle \Rightarrow \overrightarrow{a}+\overrightarrow{b}=\overrightarrow{AC}
\displaystyle \text{In } \triangle ABD
\displaystyle \overrightarrow{AD}+\overrightarrow{DB}=\overrightarrow{AB}
\displaystyle \Rightarrow \overrightarrow{b}+\overrightarrow{DB}=\overrightarrow{a}
\displaystyle \Rightarrow \overrightarrow{DB}=\overrightarrow{a}-\overrightarrow{b}
\displaystyle \text{Therefore,}
\displaystyle \overrightarrow{AC} \text{ and } \overrightarrow{DB} \text{ are the diagonals of a parallelogram whose adjacent sides are } \overrightarrow{a} \text{ and } \overrightarrow{b} \\ \text{ respectively.}

\displaystyle \textbf{Question 4: }~\text{If }\overrightarrow{a}\text{ is a vector and }m\text{ is a scalar such that }m\overrightarrow{a}=\overrightarrow{0},\text{ then what } \\ \text{are the alternatives for }m\text{ and }\overrightarrow{a}?
\displaystyle \text{Answer:}
\displaystyle \text{Given: } \overrightarrow{a} \text{ is a vector and } m \text{ is a scalar such that } m\overrightarrow{a}=\overrightarrow{0}
\displaystyle \Rightarrow \text{Either } m=0 \text{ or } \overrightarrow{a}=\overrightarrow{0}

\displaystyle \textbf{Question 5: }~\text{If }\overrightarrow{a},\overrightarrow{b}\text{ are two vectors, then write the truth value of the } \\ \text{following statements:}\\ \text{(i) }\overrightarrow{a}=-\overrightarrow{b}\Rightarrow|\overrightarrow{a}|=|\overrightarrow{b}|,\quad \text{(ii) }|\overrightarrow{a}|=|\overrightarrow{b}|\Rightarrow\overrightarrow{a}=\pm\overrightarrow{b},\quad \text{(iii) }|\overrightarrow{a}|=|\overrightarrow{b}|\Rightarrow\overrightarrow{a}=\overrightarrow{b}.
\displaystyle \text{Answer:}

\displaystyle \text{(i) }
\displaystyle \text{True.}
\displaystyle \overrightarrow{a}=-\overrightarrow{b}
\displaystyle \text{Taking modulus on both sides of the equation, we get}
\displaystyle |\overrightarrow{a}|=|-\overrightarrow{b}|
\displaystyle \Rightarrow |\overrightarrow{a}|=|\overrightarrow{b}|

\displaystyle \text{(ii) }

\displaystyle \text{False.}
\displaystyle \text{We cannot say } |\overrightarrow{a}|=|\overrightarrow{b}| \Rightarrow \overrightarrow{a}=\pm \overrightarrow{b}
\displaystyle \text{Consider an example,}
\displaystyle \overrightarrow{a}=\hat{i}+\sqrt{3}\,\hat{j} \text{ and } \overrightarrow{b}=\sqrt{2}\,\hat{i}+\sqrt{2}\,\hat{j}
\displaystyle |\overrightarrow{a}|=\sqrt{1^{2}+(\sqrt{3})^{2}}=2
\displaystyle |\overrightarrow{b}|=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}=2
\displaystyle \Rightarrow |\overrightarrow{a}|=|\overrightarrow{b}| \text{ but } \overrightarrow{a}\neq \pm \overrightarrow{b}

\displaystyle \text{(iii) }

\displaystyle \text{False.}
\displaystyle \text{We cannot say } |\overrightarrow{a}|=|\overrightarrow{b}| \Rightarrow \overrightarrow{a}=\overrightarrow{b}
\displaystyle \text{Consider an example,}
\displaystyle \overrightarrow{a}=\hat{i}+\sqrt{3}\,\hat{j} \text{ and } \overrightarrow{b}=\sqrt{2}\,\hat{i}+\sqrt{2}\,\hat{j}
\displaystyle |\overrightarrow{a}|=\sqrt{1^{2}+(\sqrt{3})^{2}}=2
\displaystyle |\overrightarrow{b}|=\sqrt{(\sqrt{2})^{2}+(\sqrt{2})^{2}}=2
\displaystyle \Rightarrow |\overrightarrow{a}|=|\overrightarrow{b}| \text{ but } \overrightarrow{a}\neq \overrightarrow{b}

\displaystyle \textbf{Question 6: }~ABCD\text{ is a quadrilateral. Find the sum of the vectors }\overrightarrow{BA},\overrightarrow{BC},\overrightarrow{CD}\text{ and }\overrightarrow{DA}.
\displaystyle \text{Answer:}
\displaystyle \text{Given: }ABCD \text{ is a quadrilateral.}
\displaystyle \text{To\ find: }\ \overrightarrow{BA}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}
\displaystyle \text{Let the position vectors of }A,B,C,D \text{ be } \overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d} \text{ respectively.}
\displaystyle \overrightarrow{BA}=\overrightarrow{a}-\overrightarrow{b}
\displaystyle \overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b}
\displaystyle \overrightarrow{CD}=\overrightarrow{d}-\overrightarrow{c}
\displaystyle \overrightarrow{DA}=\overrightarrow{a}-\overrightarrow{d}
\displaystyle \text{Therefore,}
\displaystyle \overrightarrow{BA}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA} \\ =(\overrightarrow{a}-\overrightarrow{b})+(\overrightarrow{c}-\overrightarrow{b})+(\overrightarrow{d}-\overrightarrow{c})+(\overrightarrow{a}-\overrightarrow{d})=2\overrightarrow{a}-2\overrightarrow{b}=2(\overrightarrow{a}-\overrightarrow{b})=2\overrightarrow{BA}
\displaystyle \overrightarrow{BA}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}=2\overrightarrow{BA}

\displaystyle \textbf{Question 7: }~ABCDE\text{ is a pentagon, prove that}\\ \text{(i) }\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}=\overrightarrow{0},\\ \text{(ii) }\overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}+\overrightarrow{AC}=3\overrightarrow{AC}.
\displaystyle \text{Answer:}

\displaystyle \textbf{(i)}\ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}=\overrightarrow{0}
\displaystyle \text{Let the position vectors of }A,B,C,D,E \text{ be } \overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d},\overrightarrow{e} \text{ respectively.}
\displaystyle \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a},\ \overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b},\ \overrightarrow{CD}=\overrightarrow{d}-\overrightarrow{c},\ \overrightarrow{DE}=\overrightarrow{e}-\overrightarrow{d},\ \overrightarrow{EA}=\overrightarrow{a}-\overrightarrow{e}
\displaystyle \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}=(\overrightarrow{b}-\overrightarrow{a})+(\overrightarrow{c}-\overrightarrow{b})+(\overrightarrow{d}-\overrightarrow{c})+(\overrightarrow{e}-\overrightarrow{d})+(\overrightarrow{a}-\overrightarrow{e})=\overrightarrow{0}
\displaystyle \therefore\ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA}=\overrightarrow{0}

\displaystyle \textbf{(ii)}\ \overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}+\overrightarrow{AC}=3\overrightarrow{AC}
\displaystyle \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a},\ \overrightarrow{AE}=\overrightarrow{e}-\overrightarrow{a},\ \overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b},\ \overrightarrow{DC}=\overrightarrow{c}-\overrightarrow{d},\ \overrightarrow{ED}=\overrightarrow{d}-\overrightarrow{e},\ \overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a}
\displaystyle \overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}+\overrightarrow{AC}=(\overrightarrow{b}-\overrightarrow{a})+(\overrightarrow{e}-\overrightarrow{a})+(\overrightarrow{c}-\overrightarrow{b})+(\overrightarrow{c}-\overrightarrow{d})+(\overrightarrow{d}-\overrightarrow{e})+(\overrightarrow{c}-\overrightarrow{a})=3(\overrightarrow{c}-\overrightarrow{a})=3\overrightarrow{AC}
\displaystyle \therefore\ \overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}+\overrightarrow{AC}=3\overrightarrow{AC}

\displaystyle \textbf{Question 8: }~\text{Prove that the sum of all vectors drawn from the centre of a regular } \\ \text{octagon to its vertices is the zero vector.}
\displaystyle \text{Answer:}

\displaystyle \text{Let }O\text{ be the centre of a regular octagon }ABCDEFGH.
\displaystyle \text{Then the vectors from the centre to the vertices are } \overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC},\overrightarrow{OD},\overrightarrow{OE},\overrightarrow{OF},\overrightarrow{OG},\overrightarrow{OH}.

\displaystyle \text{In a regular octagon, opposite vertices lie on the same line through the centre. Hence,}
\displaystyle \overrightarrow{OA}=-\overrightarrow{OE},\quad \overrightarrow{OB}=-\overrightarrow{OF},\quad \overrightarrow{OC}=-\overrightarrow{OG},\quad \overrightarrow{OD}=-\overrightarrow{OH}.

\displaystyle \Rightarrow \ (\overrightarrow{OA}+\overrightarrow{OE})+(\overrightarrow{OB}+\overrightarrow{OF})+(\overrightarrow{OC}+\overrightarrow{OG})+(\overrightarrow{OD}+\overrightarrow{OH})=\overrightarrow{0}.

\displaystyle \Rightarrow \ \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}+\overrightarrow{OG}+\overrightarrow{OH}=\overrightarrow{0}.

\displaystyle \therefore\ \text{The sum of all vectors drawn from the centre of a regular octagon to its vertices} \\ \text{is the zero vector.}

\displaystyle \textbf{Question 9: }~\text{If }P\text{ is a point and }ABCD\text{ is a quadrilateral and }\overrightarrow{AP}+\overrightarrow{PB}+\overrightarrow{PD}=\overrightarrow{PC},\text{ show that }ABCD\text{ is a parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: } ABCD \text{ is a quadrilateral such that } \overrightarrow{AP}+\overrightarrow{PB}+\overrightarrow{PD}=\overrightarrow{PC}
\displaystyle \text{To show: } ABCD \text{ is a parallelogram.}
\displaystyle \text{Proof: Consider,}
\displaystyle \overrightarrow{AP}+\overrightarrow{PB}+\overrightarrow{PD}=\overrightarrow{PC}
\displaystyle \Rightarrow \overrightarrow{AP}+\overrightarrow{PB}=\overrightarrow{PC}-\overrightarrow{PD}
\displaystyle \Rightarrow \overrightarrow{AB}=\overrightarrow{DC}
\displaystyle [\because \ \overrightarrow{AP}+\overrightarrow{PB}=\overrightarrow{AB} \text{ and } \overrightarrow{PD}+\overrightarrow{DC}=\overrightarrow{PC}]
\displaystyle \text{Again,}
\displaystyle \overrightarrow{AP}+\overrightarrow{PB}+\overrightarrow{PD}=\overrightarrow{PC}
\displaystyle \Rightarrow \overrightarrow{AP}+\overrightarrow{PD}=\overrightarrow{PC}-\overrightarrow{PB}
\displaystyle \Rightarrow \overrightarrow{AD}=\overrightarrow{BC}
\displaystyle [\because \ \overrightarrow{AP}+\overrightarrow{PD}=\overrightarrow{AD} \text{ and } \overrightarrow{PB}+\overrightarrow{BC}=\overrightarrow{PC}]
\displaystyle \text{Since opposite sides of the quadrilateral are equal and parallel.}
\displaystyle \text{Hence, } ABCD \text{ is a parallelogram.}

\displaystyle \textbf{Question 10: }~\text{Five forces }\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD},\overrightarrow{AE}\text{ and }\overrightarrow{AF}\text{ act at the vertex of a } \\ \text{regular hexagon }ABCDEF.\text{ Prove that the resultant is }6\overrightarrow{AO},\text{ where }O \\ \text{ is the centre of the hexagon.}
\displaystyle \text{Answer:}

\displaystyle \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}+\overrightarrow{AE}+\overrightarrow{AF}
\displaystyle \text{Consider } \triangle ADE
\displaystyle \overrightarrow{AD}+\overrightarrow{DE}+\overrightarrow{EA}=0
\displaystyle \Rightarrow \overrightarrow{AD}+\overrightarrow{DE}=\overrightarrow{AE}
\displaystyle \Rightarrow 2\overrightarrow{AO}-\overrightarrow{AB}=\overrightarrow{AE} \ [\because \ \overrightarrow{AD}=2\overrightarrow{AO} \text{ and } \overrightarrow{DE}\parallel AB \Rightarrow \overrightarrow{DE}=-\overrightarrow{AB}]
\displaystyle \Rightarrow \overrightarrow{AE}+\overrightarrow{AB}=2\overrightarrow{AO} \ \text{...(1)}
\displaystyle \text{Now, consider } \triangle ADC
\displaystyle \overrightarrow{AC}+\overrightarrow{CD}+\overrightarrow{DA}=0
\displaystyle \Rightarrow \overrightarrow{AC}+\overrightarrow{CD}=\overrightarrow{AD}
\displaystyle \Rightarrow \overrightarrow{AC}+\overrightarrow{AF}=2\overrightarrow{AO} \ [\because \ \overrightarrow{CD}=\overrightarrow{AF}] \text{...(2)}
\displaystyle \text{Using (1) and (2),}
\displaystyle \overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{AC}+\overrightarrow{AF}+\overrightarrow{AD}
\displaystyle =2\overrightarrow{AO}+2\overrightarrow{AO}+2\overrightarrow{AO}
\displaystyle =6\overrightarrow{AO}

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