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$\bar{A} = 10 \hat{i} - 10 \sqrt{3} \hat{j}$ Now $\hat{A}$ is rotated about its tail in Anti clock wise direction in xy plane by $105^{0} .$ Then the new vector $(\overset{⇀}{B})$ formed is ____ ?

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$10 \hat{i} + 10 \hat{j}$

$10 \sqrt{2} \hat{i} + 10 \sqrt{2} \hat{j}$

$10 \sqrt{2} \hat{i} - 10 \sqrt{2} \hat{j}$

$10 \sqrt{3} \hat{i} - 10 \hat{j}$

answer is C.

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Detailed Solution

$(a) \bar{A} = 10 \hat{i} - 10 \sqrt{3} \hat{j}$

$A = \sqrt{10^{2} + (- 10 \sqrt{3})^{2}} = 20$

$\tan α = \frac{10 \sqrt{3}}{10} = \sqrt{3}$

$α = 60^{\circ}$

(b) $\bar{A}$ rotated by in A.C.W direction to obtain $\bar{B}$

$\bar{B} = 20 \cos 45^{\circ} + 20 \sin 45^{\circ}$

$= 10 \sqrt{2} \hat{i} + 10 \sqrt{2} \hat{j}$

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