Question

Show that addition of vectors is commutative, that is $A + B = B + A$.

Answer 1

Hint: We will first assume vectors A and B in n dimension and then find the vector A + B and then rearrange the terms inside that to get it equal to B + A and thus we have proved.

Complete step-by-step solution:
Let us say, we have two vectors \[\overrightarrow A \] and \[\overrightarrow B \].
Let us say these are in ‘n’ dimensions.
Therefore, we can write \[\overrightarrow A \] as $ < {A_1},{A_2},{A_3},......,{A_n} > $ and \[\overrightarrow B \] as $ < {B_1},{B_2},{B_3},......,{B_n} > $.
Now, we can easily find \[\overrightarrow A + \overrightarrow B \].
\[ \Rightarrow \overrightarrow A + \overrightarrow B = < {A_1} + {B_1},{A_2} + {B_2},{A_3} + {B_3},......,{A_n} + {B_n} > \]
Now, since each of the \[{A_i}'s\] and ${B_i}'s$ are real numbers, therefore, we can write the above equation as:-
\[ \Rightarrow \overrightarrow A + \overrightarrow B = < {B_1} + {A_1},{B_2} + {A_2},{B_3} + {A_3},......,{B_n} + {A_n} > \]
This can also be termed as \[\overrightarrow B + \overrightarrow A \].
Thus, proved that vector addition is commutative.

Note: The students must note that we have used the fact that the addition of real numbers is commutative.
If we add anything which is in component form, we add the terms component – wise and thus we added the individual in vector A and vector B and reached the same conclusion.
The students must also note that we also have a parallelogram law for vector addition which requires drawing of the vectors and then adding them, eventually leading them to form a parallelogram but here we proved the commutative result for any random n – dimensional vector and reached to the required and desired result.
Let us understand in brief the parallelogram law for vector – addition:-

In the figure given above, OB gives the vector B, OA gives the vector A and since opposite sides of a parallelogram are equal, we have BC as vector A and AC as vector B as well.
Here, we can see that adding both vectors, in any form, gives us the same result and thus we have proved the parallelogram law of vector addition.
The fact we used that \[ < {B_1} + {A_1},{B_2} + {A_2},{B_3} + {A_3},......,{B_n} + {A_n} > \] results in to \[\overrightarrow B + \overrightarrow A \] has come from the definition of the vectors which helped us know that if A is any n – dimensional vector, then we can write \[\overrightarrow A \] as $ < {A_1},{A_2},{A_3},......,{A_n} > $.