Question

If `vecu` and `vecv` are vectors, then how do you proove that `∣vecu+vecv∣^2-∣vecu-vecv∣^2=4vecuvecv` ?

If `vecu` and `vecv` are vectors, then how do you proove that `∣vecu+vecv∣^2-∣vecu-vecv∣^2=4vecuvecv` ?

Answer 1

Please see a Proof in the Explanation.

Explanation:

Recall that, foe any vector `vecx, |vecx|^2=vecx*vecx`.

Therefore, for `vecu and vecv`, we have,

`|vecu+vecv|^2=(vecu+vecv)*(vecu+vecv)`.

Since dot product is distributive over vector addition, we have,

`|vecu+vecv|^2=vecu*(vecu+vecv)+vecv*(vecu+vecv)`,

`=vecu*vecu+vecu*vecv+vecv*vecu+vecv*vecv`.

But, `vecu*vecv=vecv*vecu`.

`:.|vecu+vecv|^2=|vecu|^2+vecu*vecv+vecu*vecv+|vecv|^2`.

`rArr |vecu+vecv|^2=|vecu|^2+2vecu*vecv+|vecv|^2...(star^1)`.

Similarly, `|vecu-vecv|^2=|vecu|^2-2vecu*vecv+|vecv|^2...(star^2)`.

Subtracting `(star^2)` from `(star^1)`, we immediately get,

`|vecu+vecv|^2-|vecu-vecv|^2=4vecu*vecv`.

Enjoy Maths.!