# Given $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0$, which of the following statements are correct:

Question.
Given $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0$, which of the following statements are correct:

(a) $\mathbf{a}, \mathbf{b}, \mathbf{c}$, and $\mathrm{d}$ must each be a null vector,

(b) The magnitude of $(\mathbf{a}+\mathbf{c})$ equals the magnitude of $(\mathbf{b}+\mathbf{d})$,

(c) The magnitude of a can never be greater than the sum of the magnitudes of $\mathbf{b}, \mathbf{c}$, and $\mathbf{d}$,

(d) $\mathbf{b}+\mathbf{c}$ must lie in the plane of $\mathbf{a}$ and $\mathbf{d}$ if $\mathbf{a}$ and $\mathbf{d}$ are not collinear, and in the line of $\mathbf{a}$ and $\mathbf{d}$, if they are collinear?

solution:

In order to make a + b + c + d = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.

a + b + c + d = 0

a + c = – (b + d)

Taking modulus on both the sides, we get:

| a + c | = | –(b + d)| = | b + d |

Hence, the magnitude of (a + c) is the same as the magnitude of (b + d).

a + b + c + d = 0

a = (b + c + d)

Taking modulus both sides, we get:

| a | = | b + c + d |

$|\mathbf{a}| \leq|\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}| \ldots$ (i)

Equation $(i)$ shows that the magnitude of $\mathbf{a}$ is equal to or less than the sum of the magnitudes of $\mathbf{b}, \mathbf{c}$, and $\mathbf{d}$.

Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d.