Two vectors \overrightarrow{X} and \overrightarrow{Y} have equal magnitude. The magnitude of (\overrightarrow{{X}}-{{Y}}^{\to}) is {n} times the magnitude of (\overrightarrow{{X}}+\overrightarrow{{Y}}). The angle between{{X}}^{\to} and {{Y}}^{\to} is : [25 Jul 2021 Shift 2]

Correct Answer is : cos−1(n2−1−n2−1)

\cos ^{-1}({-n^{2}-1}/{n^{2}-1})

\cos ^{-1}({n^{2}-1}/{-n^{2}-1})

\cos ^{-1}({n^{2}+1}/{-n^{2}-1})

\cos ^{-1}({n^{2}+1}/{n^{2}-1})

JEE Main 25 July 2021 Shift 2 Solved Paper

Section: Physics

Question : 20 of 90

Marks: +1, -0

Two vectors

→

and

→

have equal magnitude. The magnitude of (

→

−

→

) is n times the magnitude of (

→

). The angle between

→

and

→

is :

[25 Jul 2021 Shift 2]

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