Cho hình thoi ABCD có \(\angle BAD = {60^0}\) và BA = a. Gọi M, N lần lượt là trung điểm của AD, DC. Tính \(\overrightarrow {BM} .\overrightarrow {BN} \) bằng:

\(\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right).\overrightarrow {BC}  = {a^2}\).

\(\overrightarrow {CB} .\overrightarrow {CK}  = \dfrac{{{a^2}}}{8}\).

\(\overrightarrow {AB} .\overrightarrow {AC}  = \dfrac{{{a^2}}}{2}\).

\(\overrightarrow {CB} .\overrightarrow {CK}  = \dfrac{{{a^2}}}{2}\).

\(\overrightarrow a .\overrightarrow b  = {a_1}.{b_1} + {a_2}.{b_2}\).

$\overrightarrow a .\overrightarrow b = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|.\cos \left( {\overrightarrow a ,\overrightarrow b } \right)$.

\(\overrightarrow a .\overrightarrow b  = \dfrac{1}{2}\left[ {\overrightarrow {{a^2}}  + \overrightarrow {{b^2}}  - {{\left( {\overrightarrow a  + \overrightarrow b } \right)}^2}} \right]\).

$\overrightarrow a .\overrightarrow b = \dfrac{1}{2}\left[ {{{\left( {\overrightarrow a  + \overrightarrow b } \right)}^2} - \overrightarrow {{a^2}}  - \overrightarrow {{b^2}} } \right]$.

$30$

\(20\sqrt 2 \)

\(10\sqrt 3 \)

$20$

${45^{\rm{o}}}$.

${\rm{6}}{{\rm{0}}^{\rm{o}}}$.

${\rm{3}}{{\rm{0}}^{\rm{o}}}$.

${\rm{13}}{{\rm{5}}^{\rm{o}}}$.

\(\overrightarrow a .\overrightarrow b  = \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\).

\(\overrightarrow a .\overrightarrow b  = 0\).

\(\overrightarrow a .\overrightarrow b  =  - 1\).

\(\overrightarrow a .\overrightarrow b =  - \left| {\overrightarrow a } \right|.\left| {\overrightarrow b } \right|\).

\(\overrightarrow {BA} .\overrightarrow {BC}  = {b^2}.\)

\(\overrightarrow {BA} .\overrightarrow {BC}  = {c^2}.\)

\(\overrightarrow {BA} .\overrightarrow {BC}  = {b^2} + {c^2}.\)

\(\overrightarrow {BA} .\overrightarrow {BC}  = {b^2} - {c^2}.\)  

$2S$

$3S$

$4S$   

$6S$