Tam giác \(ABD\) có \(AB = AD = 1,\widehat {BAD} = {60^0} \Rightarrow \Delta ABD\) đều cạnh bằng \(1\)
a) \(\overrightarrow {AB} + \overrightarrow {AD} = 2\overrightarrow {AO} \Rightarrow \left| {\overrightarrow {AB} + \overrightarrow {AD} } \right| = 2\left| {\overrightarrow {AO} } \right| = 2AO = 2.AD\sin {60^0} = 2.1.\frac{{\sqrt 3 }}{2} = \sqrt 3 \)
b) \(\left| {\overrightarrow {AB} - \overrightarrow {BC} } \right| = \left| {\overrightarrow {AB} + \overrightarrow {CB} } \right| = \left| { - \overrightarrow {BA} - \overrightarrow {BC} } \right| = \left| {\overrightarrow {BA} + \overrightarrow {BC} } \right| = \left| {\overrightarrow {BD} } \right| = BD = 1\)
