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Given that vec(A)+vec(B)=vec(C ) and tha...

Given that `vec(A)+vec(B)=vec(C )` and that `vec(C )` is perpendicular to `vec(A)` Further if `|vec(A)|=|vec(C )|`, then what is the angle between `vec(A)` and `vec(B)`

A

`(pi)/(4)` radian

B

`(pi)/(2)` radian

C

`(3pi)/(4)` radian

D

`pi` radian

Text Solution

Verified by Experts

The correct Answer is:
C
`vec(C)= vec(A) + vec(B)`
Since, `vec(C) bot vec(A) and |vec(C)|= |vec(A)|`
`tan alpha= (B sin theta)/(A + B cos theta) rArr tan 90^(@)= (B sin theta)/(A+B cos theta)`
`rArr A + B cos theta= 0 rArr cos theta= - (A)/(B)`
`C^(2)= A^(2) + B^(2) + 2AB cos theta`
`rArr A^(2)= A^(2)+ B^(2) + 2AB ((- A)/(B)) rArr - B^(2)= - 2A^(2)`
`rArr A= + (1)/(sqrt2) B rArr (A)/(B)= (1)/(sqrt2)`
`cos theta= - (1)/(sqrt2)= 135^(@)= (3pi)/(4)` radian
Aliter: Since, `vec(C) bot vec(A)` [Right and triangle]
`B^(2)= A^(2) + C^(2) rArr B^(2) = 2A^(2) rArr B= sqrt2A`
Now, `A^(2) + B^(2) + 2AB cos theta= C^(2) = A^(2)`
`2A sqrt2A cos theta= - 2A^(2) rArr cos theta= - (1)/(sqrt2) rArr theta= (3pi)/(4)` rad.
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