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Q.
A man moves 20 m north, then 10 m east and then $10 \sqrt{2} m$ south-west. His displacement is
Motion in a Plane
Solution:
The situation is shown in figure.
$\overrightarrow{O A}$ represents $20 m$ north, $\overrightarrow{A B}$ represents $10 m$ east and $\overrightarrow{B C}$ represents $10 \sqrt{2} m$ south-west. If $\hat{i}$ and $\hat{j}$ be the unit vectors along east and north respectively, then
$\overrightarrow{O A}=20 \hat{j} m , \overrightarrow{A B}=10 \,\hat{i} m$
$\overrightarrow{B C} =\left(-10 \sqrt{2} \cos 45^{\circ} \hat{i}-10 \sqrt{2} \sin 45^{\circ} \hat{j}\right) m $
$=\left(-10 \sqrt{2}\left(\frac{1}{\sqrt{2}}\right) \hat{i}-10 \sqrt{2}\left(\frac{1}{\sqrt{2}}\right) \hat{j}\right) m =(-10 \hat{i}-10 \hat{j}) m$
$\therefore \quad$ Displacement $\overrightarrow{O C}=\overrightarrow{O A}+\overrightarrow{A B}+\overrightarrow{B C}$
or $\overrightarrow{O C}=20 \hat{j} m +10 \hat{i} m +(-10 \hat{i}-10 \hat{j}) m$
$ = 10 \hat{j} m = 10 m $ due north
