Question

Show that the resolution of a vector in a plane in two given directions is unique.

Answer 1

Dear student

Let assume there are two ways of resolving a vector $\overrightarrow{A}$ along two different vectors $\overrightarrow{a_1}$ and $\overrightarrow{a_2}$ such that
$$\begin{gathered} \overrightarrow{A}=\lambda \overrightarrow{a_1}+\mu \overrightarrow{a_2}={\lambda }^{\text{'}}\overrightarrow{a_1}+{\mu }^{\text{'}}\overrightarrow{a_2} \\ \Rightarrow \left(\lambda -{\lambda }^{\text{'}}\right)\overrightarrow{a_1}=\left({\mu }^{\text{'}}-\mu \right)\overrightarrow{a_2} \end{gathered}$$
where $\lambda ,\mu ,{\lambda }^{\text{'}}and{\mu }^{\text{'}}$ are real numbers.
but ​ $\overrightarrow{a_1}$ and $\overrightarrow{a_2}$ are different vectors so they can not be equal unless $\overrightarrow{a_1}=\overrightarrow{a_2}=\overrightarrow{0}$.
Hence there is only one unique way in which vector $\overrightarrow{A}$ resolve in the direction of vector $\overrightarrow{a_1}$ and $\overrightarrow{a_2}$.