Question Number 133398 by rexford last updated on 21/Feb/21
Answered by EDWIN88 last updated on 22/Feb/21
$$\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}+\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{c}}=\left(\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{c}}\right).\overset{\rightarrow} {{b}}=−\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{b}}=−\mid\overset{\rightarrow} {{b}}\mid^{\mathrm{2}} \\ $$$$\overset{\rightarrow} {{c}}.\overset{\rightarrow} {{a}}\:=−\:\mid\overset{\rightarrow} {{c}}\mid\mid\overset{\rightarrow} {{a}}\mid\left\{\frac{\mid\overset{\rightarrow} {{a}}\mid^{\mathrm{2}} +\mid\overset{\rightarrow} {{c}}\mid^{\mathrm{2}} −\mid\overset{\rightarrow} {{b}}\mid^{\mathrm{2}} }{\mathrm{2}\mid\overset{\rightarrow} {{a}}\mid\mid\overset{\rightarrow} {{c}}\mid}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\frac{\mid\overset{\rightarrow} {{b}}\mid^{\mathrm{2}} −\mid\overset{\rightarrow} {{a}}\mid^{\mathrm{2}} −\mid\overset{\rightarrow} {{c}}\mid^{\mathrm{2}} }{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Then}\:\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}+\overset{\rightarrow} {{b}}.\overset{\rightarrow} {{c}}+\overset{\rightarrow} {{c}}.\overset{\rightarrow} {{a}}=\:\frac{−\mid\overset{\rightarrow} {{a}}\mid^{\mathrm{2}} −\mid\overset{\rightarrow} {{b}}\mid^{\mathrm{2}} −\mid\overset{\rightarrow} {{c}}\mid^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 22/Feb/21
Commented by mr W last updated on 22/Feb/21
$$\boldsymbol{{a}}\centerdot\boldsymbol{{b}}=\boldsymbol{{b}}\centerdot\boldsymbol{{c}}=\boldsymbol{{c}}\centerdot\boldsymbol{{a}}=−\mathrm{1}×\mathrm{1}×\mathrm{cos}\:\mathrm{60}°=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Sigma\boldsymbol{{a}}\centerdot\boldsymbol{{b}}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$