JEE MAIN - Physics (2022 - 29th June Morning Shift - No. 12)
Two vectors $$\overrightarrow A $$ and $$\overrightarrow B $$ have equal magnitudes. If magnitude of $$\overrightarrow A $$ + $$\overrightarrow B $$ is equal to two times the magnitude of $$\overrightarrow A $$ $$-$$ $$\overrightarrow B $$, then the angle between $$\overrightarrow A $$ and $$\overrightarrow B $$ will be :
$${\sin ^{ - 1}}\left( {{3 \over 5}} \right)$$
$${\sin ^{ - 1}}\left( {{1 \over 3}} \right)$$
$${\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$
$${\cos ^{ - 1}}\left( {{1 \over 3}} \right)$$
Explanation
$$\sqrt {{A^2} + {A^2} + 2{A^2}\cos \theta } = 2\sqrt {{A^2} + {A^2} + 2{A^2}( - \cos \theta )} $$
$$ \Rightarrow 2{A^2} + 2{A^2}\cos \theta = 8{A^2} + 8{A^2} - ( - \cos \theta )$$
$$ \Rightarrow 5\cos \theta = 3$$
$$ \Rightarrow \theta = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$$
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