1.

Given : vec(a) + vec(b) + vec(c) = 0 . Out of the three vectors vec(a), vec(b) and vec(c) , two are equal in magnitude. The magnitude of the third vector is sqrt(2) times that of either of the two having equal magnitude. The angles between the vectors are :

Answer»

90°, 135°, 135°
30°, 60°, 90°
45°, 45°, 90°
45°, 60°, 90°.

Solution :Here SUPPOSE`|vec(a)|=|vec(B)|=a`
Given `veca + VECB + vecc=0`
or `veca +vecb=-vecc`
or `(veca+vecb).(veca+vecb)=(-vecc).(-vecc)`
or `a^(2)+a^(2)+2a^(2)costheta=2a^(2)`
or `vec(b)+vecc=-veca`
or same as above
or `a^(2)+2a^(2)+2(a)(sqrt(2)a)costheta=a^(2)`
or `costheta=-(2a^(2))/(2sqrt(2)a^(2))=-1/sqrt(2)`
`theta=135^@`
SIMILARLY, we can work out the third angle.


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