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Question: A vector a has components 2p and 1 with respect to a rectangular cartesian system. The system is rot...

A vector a has components 2p and 1 with respect to a rectangular cartesian system. The system is rotated through a certain angle about the origin in the anti-clockwise sense. If a has components p+1p + 1 and 1 with respect to the new system, then

A

p=0p = 0

B

p=1p = 1 or −13- \frac { 1 } { 3 }

C

p=−1p = - 1or13\frac { 1 } { 3 }

D

p=1p = 1 or −1- 1

Answer

p=1p = 1 or −13- \frac { 1 } { 3 }

Explanation

Solution

Without loss of generality, we can write

a=2pi+j=(p+1)I^+J^\mathbf { a } = 2 p \mathbf { i } + \mathbf { j } = ( p + 1 ) \hat { \mathbf { I } } + \hat { \mathbf { J } } .....(i)

Now,

J^=−sin⁡θi+cos⁡θj\hat { \mathbf { J } } = - \sin \theta \mathbf { i } + \cos \theta \mathbf { j }

∴\therefore From (i), 2pi+j=(p+1)(cos⁡θi+sin⁡θj)+(−sin⁡θi+cos⁡θj)2 p \mathbf { i } + \mathbf { j } = ( p + 1 ) ( \cos \theta \mathbf { i } + \sin \theta \mathbf { j } ) + ( - \sin \theta \mathbf { i } + \cos \theta \mathbf { j } )̃2pi+j={(p+1)cos⁡θ−sin⁡θ}i+{(p+1)sin⁡θ+cos⁡θ}j2 p \mathbf { i } + \mathbf { j } = \{ ( p + 1 ) \cos \theta - \sin \theta \} \mathbf { i } + \{ ( p + 1 ) \sin \theta + \cos \theta \} \mathbf { j }

̃ 2p=(p+1)cos⁡θ−sin⁡θ2 p = ( p + 1 ) \cos \theta - \sin \theta ....(ii) and

1=(p+1)sin⁡θ+cos⁡θ1 = ( p + 1 ) \sin \theta + \cos \theta...(iii)

Squaring and adding, 4p2+1=(p+1)2+14 p ^ { 2 } + 1 = ( p + 1 ) ^ { 2 } + 1

̃ (p+1)2=4p2( p + 1 ) ^ { 2 } = 4 p ^ { 2 } ̃ p+1=±2pp + 1 = \pm 2 p ̃ p=1p = 1, −13- \frac { 1 } { 3 }