For problem of column -I, $\overrightarrow{a}=3\stackrel{\wedge }{i}+4\stackrel{\wedge }{j},\overrightarrow{b}=2\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}-\stackrel{\wedge }{k}$ and $\overrightarrow{c}=3\stackrel{\wedge }{i}-4\stackrel{\wedge }{k}.$ Match the column – I with column – II and select the correct options.

	Column – I		Column - II
I)	$\left|\overrightarrow{a}\right|$	P)	5
II)	$|\overrightarrow{a}+\overrightarrow{b}|$	Q)	1/3
III)	$|\overrightarrow{a}-\overrightarrow{c}|$	R)	7.87
IV)	$\left|s\overrightarrow{b}\right|=1$ then ‘s’ is equal to	S)	$4\sqrt{2}$

I) $\left|\overrightarrow{a}\right|=\sqrt{3^2+4^2}=5$  
II) $\overrightarrow{a}+\overrightarrow{b}=3\stackrel{\wedge }{i}+4\stackrel{\wedge }{j}+2\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}-\stackrel{\wedge }{k}=5\stackrel{\wedge }{i}+6\stackrel{\wedge }{j}-\stackrel{\wedge }{k}$  
$\therefore |\overrightarrow{a}+\overrightarrow{b}|=\sqrt{5^2+6^2+{(-1)}^2}=\sqrt{62}=7.87$ 
III) $\overrightarrow{a}-\overrightarrow{b}=3\stackrel{\wedge }{i}+4\stackrel{\wedge }{j}-(3\stackrel{\wedge }{i}-4\stackrel{\wedge }{k})+4\stackrel{\wedge }{k}$  
$\therefore |\overrightarrow{a}-\overrightarrow{b}|=\sqrt{4^2+4^2}=4\sqrt{2}$ 
IV) $s\overrightarrow{b}=s(2\stackrel{\wedge }{i}+2\stackrel{\wedge }{j}-\stackrel{\wedge }{k})=2s\stackrel{\wedge }{i}+2s\stackrel{\wedge }{j}-s\stackrel{\wedge }{k}$ 
$\therefore \left|sb\right|=\sqrt{4s^2+4s^2+s^2}=1$ (or) $3s=1\Rightarrow s=\frac{1}{3}$ 
Hence (I) –P, (II) –R, (III) – S, (IV) –Q