If $\to u$ $vecu$ and $\to v$ $vecv$ are vectors, then how do you proove that $∣ ∣ \to u + \to v ∣^{2} - ∣ \to u - \to v ∣^{2} = 4 \to u \to v$ $∣vecu+vecv∣^2-∣vecu-vecv∣^2=4vecuvecv$ ?

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Answer 1 · 2018-06-18T16:43:31

Please see a Proof in the Explanation.

Recall that, foe any vector $\to x, {∣ ∣ \to x ∣ ∣}^{2} = \to x \cdot \to x$ $vecx, |vecx|^2=vecx*vecx$ .

Therefore, for $\to u and \to v$ $vecu and vecv$ , we have,

${∣ ∣ \to u + \to v ∣ ∣}^{2} = (\to u + \to v) \cdot (\to u + \to v)$ $|vecu+vecv|^2=(vecu+vecv)*(vecu+vecv)$ .

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

${∣ ∣ \to u + \to v ∣ ∣}^{2} = \to u \cdot (\to u + \to v) + \to v \cdot (\to u + \to v)$ $|vecu+vecv|^2=vecu*(vecu+vecv)+vecv*(vecu+vecv)$ ,

$= \to u \cdot \to u + \to u \cdot \to v + \to v \cdot \to u + \to v \cdot \to v$ $=vecu*vecu+vecu*vecv+vecv*vecu+vecv*vecv$ .

But, $\to u \cdot \to v = \to v \cdot \to u$ $vecu*vecv=vecv*vecu$ .

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

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

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

Enjoy Maths.!

If vecu→uvecu and vecv→vvecv are vectors, then how do you proove that ∣vecu+vecv∣^2-∣vecu-vecv∣^2=4vecuvecv∣∣→u+→v∣2−∣→u−→v∣2=4→u→v∣vecu+vecv∣^2-∣vecu-vecv∣^2=4vecuvecv ?