A particle undergoes three successive displacements in a plane. The first time it moves 4m south –west, the second time 5 m east and the third time 6 m in a direction $60^{\circ}$ north of east. Total displacement of particle from starting point is

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$5.17 \hat{i} - 2.37 \hat{j}$

$3.17 \hat{i} - 2.37 \hat{j}$

$5.17 \hat{i} + 2.37 \hat{j}$

$5.17 \hat{i} - 3.37 \hat{j}$

answer is C.

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Detailed Solution

$\vec{O A} = 4 \cos (180^{0} + 45^{0}) \hat{i} + 4 \sin (180^{0} + 45^{0}) \hat{j}$
$\vec{O A}$ makes $(180^{0} + 45^{0})$ angle with positive x – axis and the same with positive y-axis so
x-component of $\vec{O A} = 4 \cos (180^{0} + 45^{0})$
y- component of $\vec{O A} = 4 \sin (180^{0} + 45^{0})$
$\vec{B C} = 6 \cos 60^{0} \hat{i} + 6 \sin 60^{0} \hat{j} = 3 \hat{i} + 3 \sqrt{3} \hat{j}$ $\vec{A B} = 5 \cos 0^{0} \hat{i} = 5 \hat{i}$
$\Rightarrow \vec{O A} = - 2 \sqrt{2 \hat{i}} - 2 \sqrt{2} \hat{j}$
$\Rightarrow \vec{O C} = \vec{O A} + \vec{A B} + \vec{B C}$
$\Rightarrow \vec{O C} = (- 2 \sqrt{2} \hat{i} - 2 \sqrt{2} \hat{j}) + 5 \hat{i} + (3 \hat{i} + 3 \sqrt{3} \hat{j})$
$\Rightarrow \vec{O C} = (- 2 \sqrt{2} + 5 + 3) \hat{i} + (- 2 \sqrt{2} + 3 \sqrt{3} \hat{j})$
$\Rightarrow \vec{O C} = + 5.17 \hat{i} + 2.37 \hat{j}$