Given that `vec(A)+vec(B)+vec(C )=vec(0)`. Out of three vectors, two are equal in magnitude and the magnitude of the third vectors is `sqrt(2)` times that of either of the two having equal magnitude. Find the angles between the vectors.

Options

A. `30^(@), 60^(@), 90^(@)`
B. `45^(@), 45^(@), 90^(@)`
C. `45^(@), 60^(@), 90^(@)`
D. `90^(@), 135^(@), 135^(@)`

Answer

D

Explanation

Given A = B and `C=sqrt(2)A`
`because vecA+vecB+vecC=vec0 therefore -vecC=vecA+vecB`
`therefore |-vecC|=sqrt(A^(2)+B^(2)+2ABcostheta)`
`rArr C=sqrt(A^(2)+A^(2)+2A.A cos theta)(because B=A)`
`rArr sqrt(2)A=sqrt(2A^(2)+2A^(2)cos theta)(because C=sqrt(2)A)`
Squaring both sides
`2A^(2)=2A^(2)+2A^(2) cos theta`
`rArr 2A^(2)cos theta=0 rArr cos theta=0`
`rArr theta=90^(@)`
Since two sides are equal and `theta= 90^(@)`, therefore the triangle formed by `vecA, vecB & vecC` will be an isoceles right angled triangle.

From diagram
Angle between `vecA &vecB =90^(@)`
Angle between `vecB & vecC=180^(@)-45^(@)=135^(@)`
Angle between `vecC & vecA=180^(@)-45^(@)=135^(@)`