\displaystyle \textbf{Question 1: }\text{Find the position vector of a point }R\text{ which divides the line} \\ \text{joining the two points }P\text{ and }Q\text{ with position vectors } \overrightarrow{OP}=2\overrightarrow{a}+\overrightarrow{b}\text{ and }\overrightarrow{OQ}=\overrightarrow{a}-2\overrightarrow{b} \text{ respectively in the ratio }1:2\text{ internally and externally.}
\displaystyle \text{Answer:}
\displaystyle \text{It is given that }P\text{ and }Q\text{ are two points with position vectors } \overrightarrow{OP}=\overrightarrow{a}-2\overrightarrow{b} \text{ and } \\  \overrightarrow{OQ}=2\overrightarrow{a}+\overrightarrow{b} \text{ respectively.}
\displaystyle \text{When }R\text{ divides }PQ\text{ internally in the ratio }1:2,\text{ then}
\displaystyle \text{Position vector of }R=\frac{1(\overrightarrow{a}-2\overrightarrow{b})+2(2\overrightarrow{a}+\overrightarrow{b})}{1+2}
\displaystyle =\frac{\overrightarrow{a}-2\overrightarrow{b}+4\overrightarrow{a}+2\overrightarrow{b}}{3}
\displaystyle =\frac{5\overrightarrow{a}}{3}
\displaystyle \text{When }R\text{ divides }PQ\text{ externally in the ratio }1:2,\text{ then}
\displaystyle \text{Position vector of }R=\frac{1(\overrightarrow{a}-2\overrightarrow{b})-2(2\overrightarrow{a}+\overrightarrow{b})}{1-2}
\displaystyle =\frac{\overrightarrow{a}-2\overrightarrow{b}-4\overrightarrow{a}-2\overrightarrow{b}}{-1}
\displaystyle =\frac{-3\overrightarrow{a}-4\overrightarrow{b}}{-1}
\displaystyle =3\overrightarrow{a}+4\overrightarrow{b}

\displaystyle \textbf{Question 2: }\text{Let }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}\text{ be the position vectors of the four distinct points } \\ A,B,C,D.\text{ If }\overrightarrow{b}-\overrightarrow{a}=\overrightarrow{c}-\overrightarrow{d},\text{ then show that }ABCD\text{ is a parallelogram.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ and }\overrightarrow{d}\text{ are the position vectors of the four distinct points }A,B,C\text{ and }D\text{ respectively.}
\displaystyle \text{Also, we have }\overrightarrow{b}-\overrightarrow{a}=\overrightarrow{c}-\overrightarrow{d}.
\displaystyle \Rightarrow \overrightarrow{AB}=\overrightarrow{DC}
\displaystyle \text{Again,}
\displaystyle \overrightarrow{b}-\overrightarrow{a}=\overrightarrow{c}-\overrightarrow{d}
\displaystyle \Rightarrow \overrightarrow{b}-\overrightarrow{c}=\overrightarrow{a}-\overrightarrow{d}
\displaystyle \Rightarrow \overrightarrow{CB}=\overrightarrow{DA}
\displaystyle \text{Consequently,}
\displaystyle AB\parallel DC,\;CB\parallel DA
\displaystyle \overrightarrow{AB}=\overrightarrow{DC},\;\overrightarrow{CB}=\overrightarrow{DA}.\text{ Thus two of its opposite sides are equal and parallel.}
\displaystyle \text{Hence,}
\displaystyle ABCD\text{ is a parallelogram.}

\displaystyle \textbf{Question 3: }\text{If }\overrightarrow{a},\overrightarrow{b}\text{ are the position vectors of }A,B\text{ respectively, find the position } \\ \text{vector of a point }C\text{ in }AB\text{ produced such that }AC=3AB\text{ and that a point } \\ D\text{ in }BA\text{ produced such that }BD=2BA.
\displaystyle \text{Answer:}
\displaystyle \text{Let the position vectors of }C\text{ and }D\text{ be }\overrightarrow{c}\text{ and }\overrightarrow{d}\text{ respectively. We have,}
\displaystyle \overrightarrow{AC}=3\overrightarrow{AB}
\displaystyle \Rightarrow \overrightarrow{AC}=3(\overrightarrow{AC}-\overrightarrow{BC})
\displaystyle \Rightarrow 2\overrightarrow{AC}=3\overrightarrow{BC}
\displaystyle \Rightarrow \frac{AC}{BC}=\frac{3}{2}
\displaystyle \text{So }C\text{ divides }AB\text{ in the ratio }3:2\text{ externally.}
\displaystyle \overrightarrow{c}=\frac{2\overrightarrow{a}-3\overrightarrow{b}}{2-3}
\displaystyle =3\overrightarrow{b}-2\overrightarrow{a}
\displaystyle \text{Position vector of point }C\text{ is }3\overrightarrow{b}-2\overrightarrow{a}
\displaystyle \text{Moreover,}
\displaystyle \overrightarrow{BD}=2\overrightarrow{BA}
\displaystyle \Rightarrow \overrightarrow{BD}=2(\overrightarrow{BD}-\overrightarrow{AD})
\displaystyle \Rightarrow \overrightarrow{BD}=2\overrightarrow{AD}
\displaystyle \Rightarrow \frac{BD}{AD}=\frac{2}{1}
\displaystyle \therefore \overrightarrow{d}=\frac{\overrightarrow{b}-2\overrightarrow{a}}{1-2}
\displaystyle =2\overrightarrow{a}-\overrightarrow{b}
\displaystyle \text{Position vector of point }D\text{ is }2\overrightarrow{a}-\overrightarrow{b}

\displaystyle \textbf{Question 4: }\text{Show that the four points }A,B,C,D\text{ with position vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d} \\ \text{ respectively such that }3\overrightarrow{a}-2\overrightarrow{b}+5\overrightarrow{c}-6\overrightarrow{d}=\overrightarrow{0},\text{ are coplanar. Also, find the position} \\ \text{vector of the point of intersection of the line segments }AC\text{ and }BD.
\displaystyle \text{Answer:}
\displaystyle \text{Let }AC\text{ and }BD\text{ intersect at a point }P.\text{ We have,}
\displaystyle 3\overrightarrow{a}-2\overrightarrow{b}+5\overrightarrow{c}-6\overrightarrow{d}=\overrightarrow{0}
\displaystyle \Rightarrow 3\overrightarrow{a}+5\overrightarrow{c}=2\overrightarrow{b}+6\overrightarrow{d}
\displaystyle \text{Since the sum of coefficients on both sides is }8,
\displaystyle \text{we divide the equation on both sides by }8.
\displaystyle \Rightarrow \frac{3\overrightarrow{a}+5\overrightarrow{c}}{8}=\frac{2\overrightarrow{b}+6\overrightarrow{d}}{8}
\displaystyle \Rightarrow \frac{3\overrightarrow{a}+5\overrightarrow{c}}{3+5}=\frac{2\overrightarrow{b}+6\overrightarrow{d}}{2+6}
\displaystyle \text{Therefore, }P\text{ divides }AC\text{ in the ratio }3:5\text{ and }P\text{ divides }BD\text{ in the ratio }2:6.
\displaystyle \text{Therefore, the position vector of the point of intersection of }AC\text{ and }BD\text{ is } \frac{3\overrightarrow{a}+5\overrightarrow{c}}{8}=\frac{2\overrightarrow{b}+6\overrightarrow{d}}{8}

\displaystyle \textbf{Question 5: }\text{Show that the four points }P,Q,R,S\text{ with position vectors }\overrightarrow{p},\overrightarrow{q},\overrightarrow{r},\overrightarrow{s} \\ \text{ respectively such that }5\overrightarrow{p}-2\overrightarrow{q}+6\overrightarrow{r}-9\overrightarrow{s}=\overrightarrow{0},\text{ are coplanar. Also, find the position } \\ \text{vector of the point of intersection of the line segments }PR\text{ and }QS.
\displaystyle \text{Answer:}


\displaystyle \text{Let the point of intersection of the line segments }PR\text{ and }QS\text{ be }A.\text{ Then}
\displaystyle 5\overrightarrow{p}-2\overrightarrow{q}+6\overrightarrow{r}-9\overrightarrow{s}=\overrightarrow{0}
\displaystyle \Rightarrow 5\overrightarrow{p}+6\overrightarrow{r}=2\overrightarrow{q}+9\overrightarrow{s}
\displaystyle \text{The sum of the coefficients on both sides of the above equation is }11.
\displaystyle \text{So, we divide the given equation by }11.
\displaystyle \Rightarrow \frac{5\overrightarrow{p}+6\overrightarrow{r}}{11}=\frac{2\overrightarrow{q}+9\overrightarrow{s}}{11}
\displaystyle \Rightarrow \frac{5\overrightarrow{p}+6\overrightarrow{r}}{5+6}=\frac{2\overrightarrow{q}+9\overrightarrow{s}}{2+9}
\displaystyle \text{Therefore, }A\text{ divides }PR\text{ in the ratio }5:6\text{ and }QS\text{ in the ratio }2:9.
\displaystyle \text{The position vector of the point of intersection is } \frac{5\overrightarrow{p}+6\overrightarrow{r}}{11}=\frac{2\overrightarrow{q}+9\overrightarrow{s}}{11}

\displaystyle \textbf{Question 6: }\text{The vertices }A,B,C\text{ of triangle }ABC\text{ have respectively position vectors } \\ \overrightarrow{a},\overrightarrow{b},\overrightarrow{c} \text{ with respect to a given origin }O.\text{ Show that the point }D\text{ where the bisector} \\ \text{of }\angle A \text{ meets }BC\text{ has position vector }\overrightarrow{d}=\frac{\beta\overrightarrow{b}+\gamma\overrightarrow{c}}{\beta+\gamma},\text{ where }\beta=\lvert\overrightarrow{c}-\overrightarrow{a}\rvert\text{ and } \\ \gamma=\lvert\overrightarrow{a}-\overrightarrow{b}\rvert. \text{Hence, deduce that the incentre }I\text{ has position vector }\frac{\alpha\overrightarrow{a}+\beta\overrightarrow{b}+\gamma\overrightarrow{c}}{\alpha+\beta+\gamma}, \\ \text{ where }\alpha=\lvert\overrightarrow{b}-\overrightarrow{c}\rvert.
\displaystyle \text{Answer:}
\displaystyle \text{Let the position vectors of }A,B\text{ and }C\text{ with respect to some origin }O\text{ be }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ respectively.}
\displaystyle \text{Let }D\text{ be the point on }BC\text{ where the bisector of }\angle A\text{ meets }BC.
\displaystyle \text{Let }\overrightarrow{d}\text{ be the position vector of }D\text{ which divides }CB\text{ internally in the ratio }\beta:\gamma,\text{ where }\beta=|AC|\text{ and }\gamma=|AB|.
\displaystyle \text{Thus,}
\displaystyle \beta=|\overrightarrow{c}-\overrightarrow{a}|\text{ and }\gamma=|\overrightarrow{b}-\overrightarrow{a}|
\displaystyle \text{By section formula, the position vector of }D\text{ is given by}
\displaystyle \overrightarrow{OD}=\frac{\beta\overrightarrow{b}+\gamma\overrightarrow{c}}{\beta+\gamma}
\displaystyle \text{Let }\alpha=|\overrightarrow{b}-\overrightarrow{c}|
\displaystyle \text{Incentre is the concurrent point of angle bisectors and incentre divides the line }AD\text{ in the ratio }\alpha:(\beta+\gamma).
\displaystyle \text{So, the position vector of incentre is given as,}
\displaystyle \frac{\alpha\overrightarrow{a}+\left(\frac{\beta\overrightarrow{b}+\gamma\overrightarrow{c}}{\beta+\gamma}\right)(\beta+\gamma)}{\alpha+\beta+\gamma}
\displaystyle =\frac{\alpha\overrightarrow{a}+\beta\overrightarrow{b}+\gamma\overrightarrow{c}}{\alpha+\beta+\gamma}

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