Show that vec(a) \cdot vec(b) =frac 14|vec(a) +vec(b)|^2−frac 14|vec(a) −vec(b)|^2

waldo7852p

waldo7852p

Answered question

2022-09-21

Show that a b = 1 4 | a + b | 2 1 4 | a b | 2
I have tried:
a b = 1 4 | a + b | 2 1 4 | a b | 2 a b = 1 4 ( | a | 2 + 2 | a | | b | + | b | 2 ) 1 4 ( | a | 2 2 | a | | b | + | b | 2 ) a b = | a | | b |
However it seems that the equation only holds true when a and b are collinear.
Is that true?

Answer & Explanation

Brendon Melton

Brendon Melton

Beginner2022-09-22Added 5 answers

HINT
Such formula is known as the Polarization Identity. Due to the inner product properties, we have:
a + b 2 = a + b , a + b = a , a + 2 a , b + b , b = a 2 + 2 a , b + b 2
Analogously, we have
a b 2 = a b , a b = a , a 2 a , b + b , b = a 2 2 a , b + b 2

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