====== Question 2, Exercise 3.2 ======
Solutions of Question 2 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 2(i)=====
Find unit vector having the same direction as the vector $3\hat{i}.$
====Solution====
Let 
$$\overset{\scriptscriptstyle\rightharpoonup}{a}=3\hat{i}$$
Then
$$|\overset{\scriptscriptstyle\rightharpoonup}{a}|=\sqrt{{{(3)}^{2}}}=3$$
Now we know that
$$\hat{a}=\dfrac{{\overset{\scriptscriptstyle\rightharpoonup}{a}}}{|\overset{\scriptscriptstyle\rightharpoonup}{a}|}=\dfrac{3\hat{i}}{3}=\hat{i}$$ 
This is the required unit vector.


=====Question 2(ii)=====
Find unit vector having the same direction as the vector $3\hat{i}-4\hat{j}.$
====Solution====
Let 
$$\overset{\scriptscriptstyle\rightharpoonup}{a}=3\hat{i}-4\hat{j}$$
Then
$$|\overset{\scriptscriptstyle\rightharpoonup}{a}|=\sqrt{{{(3)}^{2}}+{{(-4)}^{2}}}=5$$
Now we know that
$$\hat{a}=\dfrac{{\overset{\scriptscriptstyle\rightharpoonup}{a}}}{|\overset{\scriptscriptstyle\rightharpoonup}{a}|}=\dfrac{3\hat{i}-4\hat{j}}{5}=\dfrac{3}{5}\hat{i}-\dfrac{4}{5}\hat{j}$$
Which is the required unit vector.

=====Question 2(iii)=====
Find unit vector having the same direction as the vector $\hat{i}+\hat{j}-2\hat{k}.$
====Solution====
Let 
$$\overset{\scriptscriptstyle\rightharpoonup}{a}=\hat{i}+\hat{j}-2\hat{k}$$ \\
Then \\
$$|\overset{\scriptscriptstyle\rightharpoonup}{a}|=\sqrt{(1)^2+(1)^2+(-2)^2}=\sqrt{6}$$ \\
Now we know that\\
$$\hat{a}=\dfrac{{\overset{\scriptscriptstyle\rightharpoonup}{a}}}{|\overset{\scriptscriptstyle\rightharpoonup}{a}|}=\dfrac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}=\dfrac{1}{\sqrt{6}}\hat{i}+\dfrac{1}{\sqrt{6}}\hat{j}-\dfrac{2}{\sqrt{6}}\hat{k}$$ \\
Which is the unit vector.

=====Question 2(iv)=====
Find unit vector having the same direction as the vector $\dfrac{\sqrt{3}}{2}\hat{i}-\dfrac{1}{2}\hat{j}.$
====Solution====
Let 
$$\overset{\scriptscriptstyle\rightharpoonup}{a}=\dfrac{\sqrt{3}}{2}\hat{i}-\dfrac{1}{2}\hat{j}$$ \\
Then \\
$$|\overset{\scriptscriptstyle\rightharpoonup}{a}|=\sqrt{{{(\dfrac{\sqrt{3}}{2})}^{2}}+{{(\dfrac{1}{2})}^{2}}}=1$$ \\
Now we know that\\
$$\hat{a}=\dfrac{{\overset{\scriptscriptstyle\rightharpoonup}{a}}}{|\overset{\scriptscriptstyle\rightharpoonup}{a}|}=\dfrac{\dfrac{\sqrt{3}}{2}\hat{i}-\dfrac{1}{2}\hat{j}}{1}=\dfrac{\sqrt{3}}{2}\hat{i}-\dfrac{1}{2}\hat{j}$$ \\
Which is the unit vector.



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