The resultant of two vectors A and B is perpendicular to vector A and its magnitude is equal to half of the magnitude of vector B . Then the angle between A and B is :
(a) `30^(@)` (b) ` 45^(@)`
(c) `150^(@)` (d) `120^(@)`

Options

A. `30^(@)`
B. `45^(@)`
C. `150^(@)`
D. `120^(@)`

Answer

Given: Resultant of `vecA&vecB` is perpendicular to Ä and magnitude of resultant `(C)=(1)/(2)vecBandalpha=90^(@)`

Explanation

(i) Magnitude of resultant:
`C^(2)=A^(2)+B^(2)+2ABcostheta`
`(B^(2))/(4)=A^(2)+B^(2)+2AB((A)/(B))`
`-(3)/(4)B^(2)=3A^(2)impliesB^(2)=4A^(2)` i.e., B=2A
(ii) Direction of resultant `tanalpha=(Bsintheta)/(A+Bcostheta)`
`oo=(Bsintheta)/(A+Bcostheta)`
`:.A+Bcostheta=0`
`costheta=-(A)/(B)=-(A)/(2A)=-(1)/(2)`
`costheta=-(1)/(2)" "theta=cos^(-1)(-(1)/(2))`
`theta=120^(@)`