(a) a+b+c+d can be zero in many ways other than a,b,c and d must each be a null vector, e.g. if the vectors are in different directions, then their resultant will be zero.
(b) Since a+b+c+d=0 ∴a+c=−(b+d)
or ∣a+c∣
(c) Since a+b+c+d=0 ∴a=−(b+c+d)
or ∣a∣=∣b+c+d∣
Therefore, the magnitude of a is equal to the magnitude of (b+c+d).
As magnitude of (b+c+d) can be equal or less than the sum of magnitudes of b,c and d but can never be greater.
Therefore, a also can never be greater than the sum of the magnitudes of b,c and d.
(d) Since a+b+c+d=0
or a+(b+c)+d=0
The resultant of three vectors a,(b+c) and d can be zero only when they lie in a plane and can be represented by the three sides of triangle taken in one order.
If a and d are collinear, then (b+c) must be in the line of a and d, only then the vector sum of all the vectors will be zero.
Thus, the statement given in option is incorrect, rest are correct.