Question

Is it possible to add two vectors of unequal magnitudes and get zero? Is it possible to add three vectors of equal magnitudes and get zero?

Answer 1

No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Example: Let us add two vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ of unequal magnitudes acting in opposite directions. The resultant vector is given by

$R=\sqrt{A^2+B^2+2AB\cos \theta }$

If two vectors are exactly opposite to each other, then
$$\begin{gathered} \theta =180°,\cos 180°=-1 \\ R=\sqrt{A^2+B^2-2AB} \\ \Rightarrow R=\sqrt{{\left(A-B\right)}^2} \\ \Rightarrow R=\left(A-B\right)\mathrm{or}\left(B-A\right) \end{gathered}$$

From the above equation, we can say that the resultant vector is zero (R = 0) when the magnitudes of the vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ are equal (A = B) and both are acting in the opposite directions.

Yes, it is possible to add three vectors of equal magnitudes and get zero.
Lets take three vectors of equal magnitudes $\overrightarrow{A,}\overrightarrow{B}\mathrm{and}\overrightarrow{C}$, given these three vectors make an angle of $120°$ with each other. Consider the figure below:


Lets examine the components of the three vectors.
$$\begin{gathered} A_x\mathit{=}A \\ {A}_y\mathit{=}0 \\ {B}_x\mathit{=}\mathit{-}B\cos 60° \\ {B}_y\mathit{=}B\sin \mathit{60}\mathit{°} \\ {C}_x\mathit{=}\mathit{-}C\cos 60° \\ {C}_y\mathit{=}\mathit{-}C\sin 60° \\ \mathrm{Here}\mathit{,}\mathit{}A\mathit{=}B\mathit{=}C \\ \mathrm{So},\mathrm{along}\mathrm{the}x-\mathrm{axis},\mathrm{we}\mathrm{have}: \\ A\mathit{-}\mathit{(}2A\cos 60°\mathit{)}\mathit{=}\mathit{0}\mathit{,}\mathit{}as\mathit{}\cos \mathit{}60°\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{} \\ \Rightarrow B\sin 60°\mathit{-}C\sin 60°\mathit{=}\mathit{0} \end{gathered}$$

Hence, proved.