====== Question 5 & 6, Exercise 3.2 ======
Solutions of Question 5 & 6 of Exercise 3.2 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

=====Question 5=====
Find the length of the vector $\overrightarrow{AB}$ from the point $\vec{A}(-3,5)$ to $\vec{B}(7,9)$. Also find the unit vector in the direction of $\overrightarrow{AB}$. 
====Solution====
The position vector of $\vec{A}$ and $\vec{B}$ are
$$\overrightarrow{OA}=-3\hat{i}+5\hat{j},$$ 
$$\overrightarrow{OB}=7\hat{i}+9\hat{j}.$$
Thus we have
\begin{align}\overrightarrow{AB}&=\overrightarrow{OB}-\overrightarrow{OA} \\
&=7\hat{i}+9\hat{j}-(-3\hat{i}+5\hat{j}) \\
&=10\hat{i}+4\hat{j}.\end{align}
This gives
\begin{align}|\overrightarrow{AB}|&=\sqrt{(10)^2+(4)^2}\\
&=\sqrt{116} = 2\sqrt{29}.\end{align}
Let $\hat{r}$ be unit vector in the direction of $\overrightarrow{AB}$. Then
\begin{align}\hat{r}&=\dfrac{\overrightarrow{AB}}{|\overrightarrow{AB}|}\\
&=\dfrac{10\hat{i}+4\hat{j}}{2\sqrt{29}} \\
&=\dfrac{5}{\sqrt{29}}\hat{i}+\dfrac{2}{\sqrt{29}}\hat{j}. \end{align}
Hence length of $\overrightarrow{AB}$ is $2\sqrt{29}$ and the unit vector in the direction of $\overrightarrow{AB}$ is $\dfrac{5}{\sqrt{29}}\hat{i}+\dfrac{2}{\sqrt{29}}\hat{j}$.

=====Question 6=====
If $ABCD$ is the parallelogram such that the coordinates of the vertices $A, B$ and $C$ respectively given by $(-2,-3),(1,4)$ and $(0,5).$ find coordinates of the vertex $D.$
====Solution====
Position vectors of given points are
$$\overrightarrow{OA}=-2\hat{i}-3\hat{j},$$
$$\overrightarrow{OB}=\hat{i}+4\hat{j},$$ 
$$\overrightarrow{OC}=5\hat{j}.$$
Let $\overrightarrow{OD}=x\hat{i}+y\hat{j}$. Since $ABCD$ is parallelogram, we have that
$$\overrightarrow{AB}=\overrightarrow{DC}.$$
This gives
\begin{align}&\overrightarrow{OB}-\overrightarrow{OA}=\overrightarrow{OC}-\overrightarrow{OD}\\
\implies &(\hat{i}+4\hat{j})-(-2\hat{i}-3\hat{j})=5\hat{j}-x\hat{i}-y\hat{j}\\ 
\implies &3\hat{i}+7\hat{j}=-x\hat{i}+(5-y)\hat{j}\end{align}
By comparison, we have
$$3=-x \text{ and } 7=5-y$$
This gives
$$x=-3 \text{ and } y=-2.$$
Hence coordinates of $D$ are $(-3,-2)$.   

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