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JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Vector

Two vectors

A

and

B

are defined as

A = a \overset{}{^}

and

B = a (cos ω t \overset{}{^} + sin ω t \overset{}{^})

, where

a

is a constant and

ω = π /6 r a d s^{- 1}

. If

∣ A + B ∣ = 3 ∣ A - B ∣

at time

t = τ

for the first time, the value of

τ

, in seconds, is ____.

Answer: 2

Solution

A + B = a (1 + cos ω t) \hat{i} + a sin ω t \hat{j}

and

A - B = a (1 - cos ω t) \hat{i} - a sin ω t \hat{j}

. The condition

∣ A + B ∣ = 3 ∣ A - B ∣

gives:

a^{2} [(1 + cos ω t)^{2} + sin^{2} ω t] = 3 a^{2} [(1 - cos ω t)^{2} + sin^{2} ω t]

. Simplifying:

2 + 2 cos ω t = 6 - 6 cos ω t

, so

cos ω t = 1/2

. For the first time,

ω t = π /3

, giving

t = \frac{π /3}{π /6} = 2

seconds.

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer

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Three vectors $P, Q$ and $R$ are shown in the figure. Let $S$ be any point on the vector $R$ . The distance between the points $P$ and $S$ is $b ∣ R ∣$ . The general relation among vectors $P, Q$ and $S$ is
JEE Advanced 2017View solution →