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If $\vec{A}=4 \hat{i}-3 \hat{j}$ and $\vec{B}=6 \hat{i}+8 \hat{j}$ then magnitude and direction of $\dot{A}+\dot{B}$ will be
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The correct answer is:
$5 \sqrt{5}, \tan ^{-1}(1 / 2)$
$\vec{A}+\vec{B}=4 \hat{i}-3 \hat{j}+6 \hat{i}+8 \hat{j}=10 \hat{i}+5 \hat{j}$
$|\vec{A}+\vec{B}|=\sqrt{(10)^2+(5)^2}=5 \sqrt{5}$
$\tan \theta=\frac{5}{10}=\frac{1}{2} \Rightarrow \theta=\tan ^{-1}\left(\frac{1}{2}\right)$
$|\vec{A}+\vec{B}|=\sqrt{(10)^2+(5)^2}=5 \sqrt{5}$
$\tan \theta=\frac{5}{10}=\frac{1}{2} \Rightarrow \theta=\tan ^{-1}\left(\frac{1}{2}\right)$
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