Cho tam giác \(ABC\) cân tại \(A\), \(\widehat A = {120^0}\) và \(AB = a\). Tính \(\overrightarrow {BA} .\overrightarrow {CA} \)

\(2{a^2}\)

\({a^2}\)  

\({a^2}\sqrt 2 \)

0

\(\alpha  = {180^0}.\)

\(\alpha  = {0^0}.\)

\(\alpha  = {90^0}.\)

\(\alpha  = {45^0}.\)

\(\overrightarrow {AH} .\overrightarrow {BC}  = 0.\)

\(\left( {\overrightarrow {AB} ,\overrightarrow {HA} } \right) = {150^0}.\)

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

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

\(\overrightarrow {AB} .\overrightarrow {AC}  = \dfrac{1}{2}{a^2}.\)

\(\overrightarrow {AC} .\overrightarrow {CB}  =  - \dfrac{1}{2}{a^2}.\)

\(\overrightarrow {GA} .\overrightarrow {GB}  = \dfrac{{{a^2}}}{6}.\)

\(\overrightarrow {AB} .\overrightarrow {AG}  = \dfrac{1}{2}{a^2}.\)

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

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

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

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

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

\(\overrightarrow {AE} .\overrightarrow {AB}  = \sqrt 3 {a^2}.\)

\(\overrightarrow {AE} .\overrightarrow {AB}  = \sqrt 5 {a^2}.\)

\(\overrightarrow {AE} .\overrightarrow {AB}  = 5{a^2}.\)

\(P = 2\sqrt 2 a.\)

\(P = 2{a^2}.\)         

\(P = {a^2}.\)

\(P =  - 2{a^2}.\)