Question Number 170079 by depressiveshrek last updated on 16/May/22
$$\mid\overset{\rightarrow} {{a}}\mid=\mathrm{1} \\ $$$$\mid\overset{\rightarrow} {{b}}\mid=\mathrm{2} \\ $$$$\sphericalangle\left(\overset{\rightarrow} {{a}},\:\overset{\rightarrow} {{b}}\right)=\frac{\pi}{\mathrm{3}} \\ $$$$\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}\mid=? \\ $$
Answered by som(math1967) last updated on 16/May/22
$$\:\overset{\rightarrow} {{a}}.\overset{\rightarrow} {{b}}=\mathrm{1}.\mathrm{2}.{cos}\frac{\pi}{\mathrm{3}}=\mathrm{1} \\ $$$$\mid\overset{\rightarrow} {{a}}\:+\overset{\rightarrow} {{b}}\mid=\sqrt{\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{2}.\mathrm{1}}=\sqrt{\mathrm{7}} \\ $$
Answered by mr W last updated on 16/May/22
Commented by mr W last updated on 16/May/22
$$\mid{a}+{b}\mid^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}×\mathrm{2}\:\mathrm{cos}\:\left(\pi−\frac{\pi}{\mathrm{3}}\right) \\ $$$$\mid{a}+{b}\mid^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{2}×\mathrm{1}×\mathrm{2}\:\mathrm{cos}\:\left(\frac{\pi}{\mathrm{3}}\right) \\ $$$$\mid{a}+{b}\mid^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{2}×\mathrm{1}×\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mid{a}+{b}\mid^{\mathrm{2}} =\mathrm{7} \\ $$$$\mid{a}+{b}\mid=\sqrt{\mathrm{7}} \\ $$