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

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Answer 1 · 2018-06-18T16:43:31

Please see a Proof in the 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, $vecu*vecv=vecv*vecu$ .

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

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

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

Enjoy Maths.!

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