Vectors basics

$\text{Part 1}$

The resultant of two vectors $A$ and $B$ with angle $\theta$ between them is given by:

$\begin{aligned} |R| &= \sqrt{A^2+B^2+2AB \cos \theta} \\ \text{Given } &\rightarrow A=1, B=1,R=1 \\ \implies 1 &= \sqrt{1+1+2\cos \theta} \\ \text{Squaring both sides, } 1&= 1+1+2\cos \theta \\ \implies \cos \theta &= -\dfrac{1}{2}\\ \implies \theta &= \boxed{120^{\circ} }\\ \end{aligned}$

$\text{Part 2}$

Two vectors of equal magnitude, but opposite directions, sum to $0$ . But since, we are not allowed to have vectors of equal magnitude, we need at least $\boxed{3}$ vectors. The three vectors forming a closed triangle, by the Triangle Law of Vector Addition, will have a sum of $0$ .

$\text{Answer}$

$\boxed{120+3=123}$

(i) Let us consider three vectors $\vec a, \vec b, \vec c$ of equal magnitude : $|\vec a|=|\vec b|=|\vec c |=m$ , such that $\vec c =\vec a+\vec b$

Then, $|\vec c |=|\vec a +\vec b|$

$\implies m=\sqrt {m^2+m^2+2m^2\cos α}=2m\cos (\frac{α}{2})$ ,

where $α$ is the angle between $\vec a$ and $\vec b$

$\implies \cos (\frac{α}{2})=\frac 12\implies α=120\degree$

This holds true for any three vectors of equal magnitude. Since unit vectors have the same magnitude $1$ , this holds for them as well.

(ii) Since $||\vec a |-|\vec b ||\leq |\vec a +\vec b |\leq |\vec a |+|\vec b |$ ,

therefore two vectors of unequal magnitudes can never give a zero sum

Using vector addition algorithm, we get that sum of a number of vectors is zero if they can be represented by the sides of a polygon taken in order. Since a polygon of non-zero area has a minimum of three sides, the minimum number of such vectors is $3$

Hence the answer is $120+3=\boxed {123}$ .

Here @Vinayak Srivastava

Try this problem, I think it's within your reach for sure:

https://brilliant.org/problems/the-classic-rope-on-table-problem-with-a-twist/?ref_id=1600873