Physics · 2019
JEE Main · 10 January 2019, Shift 2 · Q2
Two vectors A and B have equal magnitudes. The magnitude of ( A + B ) is ' n ' times the magnitude of ( A - B ). The angle between A and B is:
Two vectors $\displaystyle \vec{A}$ and $\displaystyle \vec{B}$ have equal magnitudes. The magnitude of $\displaystyle (\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})$ is ' $\displaystyle n$ ' times the magnitude of $\displaystyle (\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}})$. The angle between $\displaystyle \vec{A}$ and $\displaystyle \vec{B}$ is :
Official answer
(1)
$\displaystyle \cos ^{-1}\left[\frac{\mathrm{n}^2-1}{\mathrm{n}^2+1}\right]$
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JEE Main 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.