`text{∆ABC vuông tại A có AB=a; BC=2a}`
Ta có:
`sinACB= {AB}/{BC}=a/{2a}= 1/ 2`
`=>\hat{ACB}=30°`
`=>\hat{ABC}=90°-30°=60°`
`AC^2=BC^2-AB^2=(2a)^2-a^2=3a^2`
`=>AC=a\sqrt{3}`
*`\vec{CA}.\vec{CB}=CA.CB.cosACB`
`=a\sqrt{3}.2a.cos30°=2a^2 \sqrt{3}. {\sqrt{3}}/2=3a^2`
*`\vec{BA}.\vec{BC}=BA.BC.cosABC`
`=a.2a.cos60°=2a^2. 1/2=a^2`
*`\vec{AC}.\vec{CB}=-\vec{CA}.\vec{CB}=-3a^2`
Vậy:
`\vec{CA}.\vec{CB}=3a^2`
`\vec{BA}.\vec{BC}=a^2`
`\vec{AC}.\vec{CB}=-3a^2`
