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I need some help on this question. I just have no idea on how to get started on this problem.

Here is the problem:

For two and three dimensional vectors, the fundamental property of dot products

$ A \cdot B = |A||B| \cos{\theta}$ implies that $A \cdot B \leq |A||B|$ (1.1)

Show that $|A- \gamma B|^2$ implies (1.1) where $\gamma = \frac{A \cdot B}{B \cdot B}$

I just do not know how to begin with this problem

  • I believe you're missing something in your "show", as you the part before the implication is just the square of a term....not a square of a term being equal to something, or geater than something, or.... – Alan Sep 30 '14 at 01:57
  • I think I need to show that $|A-\gamma B|^2 \leq |A|^2 - |\gamma B|^2$ if that makes any sense –  Sep 30 '14 at 02:14

1 Answers1

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If $|B|=0$, then $|A||B|\geq|A\cdot B|$ holds trivially. Else, consider: $$ 0\leq |A-\gamma B|^2=|A|^2-2\gamma A\cdot B+\gamma^2|B|^2. $$ Now, if you use $\gamma=\frac{A\cdot B}{B\cdot B}$, then the above is translated to $$ 0\leq |A|^2-2\frac{(A\cdot B)^2}{|B|^2}+\frac{(A\cdot B)^2}{|B|^2}=|A|^2-\frac{(A\cdot B)^2}{|B|^2}\implies |A|^2\geq \frac{(A\cdot B)^2}{|B|^2} $$ which rearranges to give $|A|^2|B|^2\geq(A\cdot B)^2$, or, equivalently, $|A||B|\geq|A\cdot B|$.

Kim Jong Un
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  • Okay going back to when you set $\gamma = \frac{A \cdot B}{B \cdot B}$, would the equation turn out to be $0 \leq |A|^2 -2\frac{(A \cdot B)^2}{|B|^2} +\frac{(A \cdot B)}{|B|^2} |B|^2$ –  Oct 01 '14 at 01:58
  • @ SSivetts: the last term is $\gamma^2|B|^2$. You can maybe check your calculation. – Kim Jong Un Oct 01 '14 at 01:59
  • Yeah, I got that, but it turns out I have $0 \leq |A|^2 -2\frac{(A \cdot B)^2}{|B|^2} +\frac{(A \cdot B)^2}{|B|^2} |B|^2$ However, in your work, the $|B|^2$ term does not show up. Why is that? –  Oct 01 '14 at 02:03
  • Note that $\gamma=\frac{A\cdot B}{B\cdot B}=\frac{A\cdot B}{|B|^2}$ so $\gamma^2=\frac{(A\cdot B)^2}{|B|^\color{red}{4}}$. Now multiply by $|B|^2$. – Kim Jong Un Oct 01 '14 at 02:05
  • Oh I got it now. Thanks a lot for your help and patience. I really appreciate it. –  Oct 01 '14 at 02:06