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Question-216983




Question Number 216983 by hardmath last updated on 26/Feb/25
Answered by mr W last updated on 26/Feb/25
A^→ =(a_1 ,a_2 ,..,a_n )  B^→ =(b_1 ,b_2 ,..,b_n )  A^→ ∙B^(→) =∣A∣∣B∣ cos θ≤∣A∣∣B∣  ⇒a_1 b_1 +a_2 b_2 +...+a_n b_n ≤(√(a_1 ^2 +a_2 ^2 +...+a_n ^2 ))∙(√(b_1 ^2 +b_2 ^2 +...+b_n ^2 ))
$$\overset{\rightarrow} {\boldsymbol{{A}}}=\left({a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,..,{a}_{{n}} \right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{B}}}=\left({b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,..,{b}_{{n}} \right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{A}}}\centerdot\overset{\rightarrow} {\boldsymbol{{B}}}=\mid\boldsymbol{{A}}\mid\mid\boldsymbol{{B}}\mid\:\mathrm{cos}\:\theta\leqslant\mid\boldsymbol{{A}}\mid\mid\boldsymbol{{B}}\mid \\ $$$$\Rightarrow{a}_{\mathrm{1}} {b}_{\mathrm{1}} +{a}_{\mathrm{2}} {b}_{\mathrm{2}} +…+{a}_{{n}} {b}_{{n}} \leqslant\sqrt{{a}_{\mathrm{1}} ^{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} +…+{a}_{{n}} ^{\mathrm{2}} }\centerdot\sqrt{{b}_{\mathrm{1}} ^{\mathrm{2}} +{b}_{\mathrm{2}} ^{\mathrm{2}} +…+{b}_{{n}} ^{\mathrm{2}} } \\ $$
Commented by mr W last updated on 26/Feb/25
the scalar product of two vectors
$${the}\:{scalar}\:{product}\:{of}\:{two}\:{vectors} \\ $$
Commented by hardmath last updated on 26/Feb/25
  Dear professor, what is the basic proof of this?
$$ \\ $$Dear professor, what is the basic proof of this?
Commented by mr W last updated on 26/Feb/25
https://en.m.wikipedia.org/wiki/Dot_product

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