Explanation

(a) $\overrightarrow{A} = 2\widehat{i} + 4\widehat{j},$ $\overrightarrow{B} = 5\widehat{i} + p\ \widehat{j}$
$A = \sqrt{2^{2} + 4^{2}} = \sqrt{20}$
$B = \sqrt{5^{2} + p^{2}}$
$\overrightarrow{A} \bullet \overrightarrow{B} = 10 - 4p$
If $\overrightarrow{A} \parallel \overrightarrow{B}$ then
$\overrightarrow{A} \bullet \overrightarrow{B} = AB\ \cos 0{^\circ} = AB$
$10 - 4p = \sqrt{20}\sqrt{25 + p^{2}}$
Square 100+16$p^{2} - 80p$
$= 20\left( 25 + p^{2} \right) = 500 = 20p^{2}$
or $20p^{2} - 16p^{2} + 80p + 400 = 0$
or $p^{2} + 20p + 100 = 0$
or $(p + 10)^{2} = 0$
$\therefore\ \ \ \ \ \ \ \ \ p = - 10$
$\therefore\ \ \ \ \ \overrightarrow{B} = \ \ 5\widehat{i} + 10\widehat{j}$
$B = \sqrt{5^{2} + (10)^{2}} = \sqrt{125} = 5\sqrt{5}$