## In This post you will get answer of Q.

Given $vec{A} + vec{B} + vec{C} + vec{D} = vec{0}$, which of the following statements is not correct ?

Solution:

(a) The statement is not correct. It is because $vec{A} + vec{B} + vec{C} + vec{D}$ can be zero in many ways other than $vec{A}$, $vec{B}$, $vec{C}$ and $vec{D}$ being each a null vector.

(b) The statement is correct as proved below

$vec{A}+vec{B}+vec{C}+vec{D}=vec{0}$

or $vec{A}+vec{C}=-left(vec{B}+vec{D}right)$

$therefore left|vec{A}+vec{C}right|=left|vec{B}+vec{D}right|$

(c) The statement is correct and can be proved using triangular inequality.

$left|vec{B}right|+left|vec{C}right| geleft|vec{R}right|,…left(iright)$

$left|vec{R}right|+left|vec{D}right| ge left|vec{A}right|,…left(iiright)$

Using (i) and (ii),

$left|vec{A}right|leleft|vec{B}right| +left|vec{C}right|+left|vec{D}right|$

(d) The statement is correct as proved below

$vec{A} + vec{B} + vec{C} + vec{D} = 0$

or $vec{A} + (vec{B} + vec{C}) + vec{D} = 0$

The resultant sum of three vectors $vec{A}$, $(vec{B} + vec{C})$ and $vec{D}$ can be zero only if $(vec{B} + vec{C})$ lies in the plane of $vec{A}$ and $vec{D}$ and these three vectors are represented by the three sides of a triangle taken in one order. If $vec{A}$ and $vec{D}$ are collinear, then $(vec{B} + vec{C})$ must be in the line of $vec{A}$ and $vec{D}$, only then the vector sum of all the vectors will be zero.