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given that vecA+vecB+vecC=0 out of three...

given that `vecA+vecB+vecC=0` out of three vectors two are equal in magnitude and the magnitude of third vector is `sqrt2` times that of either of the two having equal magnitude. Then the angles between vectors are given by

A

`30^(@), 60^(@), 90^(@)`

B

`45^(@), 45^(@), 90^(@)`

C

`45^(@), 60^(@), 90^(@)`

D

`90^(@), 135^(@), 135^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
`vec(A) + vec(B) + vec(C)= 0`
From polygon law, three vectors having summation zero should form a closed polygon (Triangle)
`|vec(A)| = |vec(B)| and |vec(C)| = sqrt2 |vec(A)|`

Angle between A and B `alpha= 90^(@)`
Angle between B and C, `beta= 135^(@)`
Angle between A and C, `gamma= 135^(@)`
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