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

Concept: Triangle Inequality — the magnitude of a sum of vectors is at most the sum of their magnitudes, and at least the absolute difference.

Step 1 — Statement (a):

The condition $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0$ does not force each vector to be null. Example: $\vec{a} = (1,0), \vec{b} = (-1,0), \vec{c} = (0,1), \vec{d} = (0,-1)$ satisfies the sum but none is null.

So (a) is incorrect.

Step 2 — Statement (b):

From $\vec{a} + \vec{b} + \vec{c} + \vec{d} = 0$, we have $\vec{a} + \vec{c} = -(\vec{b} + \vec{d})$. Magnitudes of a vector and its negative are equal.

So (b) is correct.

Step 3 — Statement (c):

By the triangle inequality: $|\vec{a}| = | -(\vec{b} + \vec{c} + \vec{d}) | \le |\vec{b}| + |\vec{c}| + |\vec{d}|$.

So (c) is correct.

Step 4 — Statement (d): …

The key idea is the Triangle Inequality: for any two vectors, $|\vec{x}+\vec{y}| \le |\vec{x}|+|\vec{y}|$, with equality only when they point in the same direction. This inequality extends naturally to more vectors and is the backbone of checking magnitude claims. Also, the zero-sum condition means the four vectors form a closed quadrilateral when placed head-to-tail — a geometric picture that helps with direction-based statements.

Let’s examine each statement one by one.

1. Statement (a): “$\vec{a}$, $\vec{b}$, $\vec{c}$, and $\vec{d}$ must each be a null vector.”

This is false. A simple counterexample: take $\vec{a} = (1,0)$, $\vec{b} = (-1,0)$, $\vec{c} = (2,0)$, $\vec{d} = (-2,0)$. Their sum is zero, yet none is a null vector. The condition only says the total sum vanishes; individual vectors can be non-zero as long as they cancel out.

2. Statement (b): “The magnitude of $(\vec{a}+\vec{c})$ equals the magnitude of $(\vec{b}+\vec{d})$.”

From $\vec{a}+\vec{b}+\vec{c}+\vec{d}=0$, we can rearrange:

Taking magnitudes on both sides:

So the magnitudes are always equal. This is a direct algebraic consequence — no extra conditions needed. Statement (b) is correct.

3. Statement (c): “The magnitude of $\vec{a}$ can never be greater than the sum of the magnitudes of $\vec{b}$, $\vec{c}$, and $\vec{d}$.”

From the given equation, $\vec{a} = -(\vec{b}+\vec{c}+\vec{d})$. By the Triangle Inequality:

So $|\vec{a}|$ is always less than or equal to that sum — it can never exceed it. Statement (c) is correct. …

Concept: A Closed Polygon, Reordered as Needed

Method: The Closed-Quadrilateral Picture, Using Commutativity to Re-Pair Vectors as Diagonals

Since $\vec a+\vec b+\vec c+\vec d=\vec 0$, placing the four vectors head-to-tail in any order returns to the starting point -- because vector addition is commutative, every reordering of $\vec a,\vec b,\vec c,\vec d$ closes into a valid quadrilateral. This method exploits that freedom: for each statement, the vectors are placed in whichever order makes the quantities being compared into the two diagonals of that particular closed shape.

Statement (a): must each vector be null?

Build an explicit non-degenerate closed quadrilateral: $\vec a=(1,0)$, $\vec b=(0,1)$, $\vec c=(-1,0)$, $\vec d=(0,-1)$ -- a unit square, traversed corner to corner. These sum to $\vec 0$ (the square closes), yet none of the four sides is a null vector.

Verdict: False.

Statement (b): $|\vec a+\vec c| = |\vec b+\vec d|$?

Reorder the same four vectors into the sequence $\vec a,\vec c,\vec b,\vec d$ (legal, since addition is commutative -- $\vec a+\vec c+\vec b+\vec d=\vec 0$ too). Placed head-to-tail in this order, they form a new closed quadrilateral $P_0P_1P_2P_3$ with sides $\vec a,\vec c,\vec b,\vec d$. The diagonal $P_0P_2 = \vec a+\vec c$ (the first two sides); and because the whole figure closes, the remaining two sides get you back the same distance the other way round: $P_2P_0 = \vec b+\vec d = -(\vec a+\vec c)$. A diagonal traversed in either direction has the same length:

Verdict: True.

Statement (c): can $|\vec a|$ exceed $|\vec b|+|\vec c|+|\vec d|$?

In the original ordering $\vec a,\vec b,\vec c,\vec d$ (closed quadrilateral $Q_0Q_1Q_2Q_3$, $\vec a=Q_0Q_1$), the sum $\vec b+\vec c+\vec d$ is the "3-side detour" from $Q_1$ all the way around to $Q_0$ (via $Q_2,Q_3$) -- while $-\vec a$ is the direct single segment from $Q_1$ back to $Q_0$. A detour made of several straight legs between two points can never be shorter than the direct segment between them (repeated triangle inequality) -- it can only be longer or, in the degenerate straight-line case, equal:

Verdict: True (can never exceed).

Statement (d): must $\vec b+\vec c$ lie in the plane/line of $\vec a,\vec d$? …