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← JEE Advanced 2018 · Paper 1
PhysicsNumerical+3Question 7 of 54

JEE Advanced 2018 · Paper 1, Physics question 7

Two vectors A⃗\vec A and B⃗\vec B are defined as A⃗=a i^\vec A = a\,\hat i and B⃗=a(cos⁡ωt i^+sin⁡ωt j^)\vec B = a(\cos\omega t\,\hat i + \sin\omega t\,\hat j), where aa is a constant and ω=π/6 rad s−1\omega = \pi/6\ \mathrm{rad\,s^{-1}}. If ∣A⃗+B⃗∣=3∣A⃗−B⃗∣|\vec A + \vec B| = \sqrt3|\vec A - \vec B| at time t=τt = \tau for the first time, the value of τ\tau, in seconds, is ______.

Answer and worked solution

Official answer: 2

Both vectors have magnitude aa, and the angle between them is θ=ωt\theta=\omega t.
∣A⃗+B⃗∣=2acos⁡θ2,∣A⃗−B⃗∣=2asin⁡θ2.|\vec A+\vec B|=2a\cos\frac\theta2,\qquad|\vec A-\vec B|=2a\sin\frac\theta2.

Condition.
cot⁡θ2=3  ⟹  θ2=π6  ⟹  θ=π3.\cot\frac\theta2=\sqrt3\implies\frac\theta2=\frac\pi6\implies\theta=\frac\pi3.

Time. π6τ=π3  ⟹  τ=2\dfrac\pi6\tau=\dfrac\pi3\implies\tau=2 s.

Answer: 2

Topics:Vectors