waldo7852p

Answered

2022-09-21

Show that $\stackrel{\to }{a}\cdot \stackrel{\to }{b}=\frac{1}{4}|\stackrel{\to }{a}+\stackrel{\to }{b}{|}^{2}-\frac{1}{4}|\stackrel{\to }{a}-\stackrel{\to }{b}{|}^{2}$
I have tried:
$\stackrel{\to }{a}\cdot \stackrel{\to }{b}=\frac{1}{4}|\stackrel{\to }{a}+\stackrel{\to }{b}{|}^{2}-\frac{1}{4}|\stackrel{\to }{a}-\stackrel{\to }{b}{|}^{2}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}\cdot \stackrel{\to }{b}=\frac{1}{4}\left(|\stackrel{\to }{a}{|}^{2}+2|\stackrel{\to }{a}||\stackrel{\to }{b}|+|\stackrel{\to }{b}{|}^{2}\right)-\frac{1}{4}\left(|\stackrel{\to }{a}{|}^{2}-2|\stackrel{\to }{a}||\stackrel{\to }{b}|+|\stackrel{\to }{b}{|}^{2}\right)\phantom{\rule{0ex}{0ex}}\stackrel{\to }{a}\cdot \stackrel{\to }{b}=|\stackrel{\to }{a}||\stackrel{\to }{b}|$
However it seems that the equation only holds true when $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$ are collinear.
Is that true?

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Answer & Explanation

Brendon Melton

Expert

2022-09-22Added 5 answers

HINT
Such formula is known as the Polarization Identity. Due to the inner product properties, we have:
$\begin{array}{r}‖\mathbf{\text{a}}+\mathbf{\text{b}}{‖}^{2}=⟨\mathbf{\text{a}}+\mathbf{\text{b}},\mathbf{\text{a}}+\mathbf{\text{b}}⟩=⟨\mathbf{\text{a}},\mathbf{\text{a}}⟩+2⟨\mathbf{\text{a}},\mathbf{\text{b}}⟩+⟨\mathbf{\text{b}},\mathbf{\text{b}}⟩=‖\mathbf{\text{a}}{‖}^{2}+2⟨\mathbf{\text{a}},\mathbf{\text{b}}⟩+‖\mathbf{\text{b}}{‖}^{2}\end{array}$
Analogously, we have
$\begin{array}{r}‖\mathbf{\text{a}}-\mathbf{\text{b}}{‖}^{2}=⟨\mathbf{\text{a}}-\mathbf{\text{b}},\mathbf{\text{a}}-\mathbf{\text{b}}⟩=⟨\mathbf{\text{a}},\mathbf{\text{a}}⟩-2⟨\mathbf{\text{a}},\mathbf{\text{b}}⟩+⟨\mathbf{\text{b}},\mathbf{\text{b}}⟩=‖\mathbf{\text{a}}{‖}^{2}-2⟨\mathbf{\text{a}},\mathbf{\text{b}}⟩+‖\mathbf{\text{b}}{‖}^{2}\end{array}$

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