MHB Find $\vec{AC}.\vec{AB}$ in Rectangle $ABCD$

[Albert1]

Rectangle $ABCD$ given :$\angle CAB=30^o$ ,and $\vec{AC}.\vec{AD}=\mid\vec{AC}\,\, \mid$

please find the value of :$\vec{AC}.\vec{AB}$

 

[Greg]

Spoiler

$$\vec{AC}\cdot\vec{AD}=|\vec{AC}||\vec{AD}|\cos(60)=\frac12|\vec{AC}||\vec{AD}|=|\vec{AC}|\implies|\vec{AD}|=2$$

$$|\vec{AC}|\cos(60)=2\implies|\vec{AC}|=4\implies|\vec{AB}|=\sqrt{12}$$

$$\vec{AC}\cdot\vec{AB}=4\sqrt{12}\cos(30)=12$$

 

[Albert1]

Spoiler

$$\vec{AC}\cdot\vec{AD}=|\vec{AC}||\vec{AD}|\cos(60)=\frac12|\vec{AC}||\vec{AD}|=|\vec{AC}|\implies|\vec{AD}|=2$$

$$|\vec{AC}|\cos(60)=2\implies|\vec{AC}|=4\implies|\vec{AB}|=\sqrt{12}$$

$$\vec{AC}\cdot\vec{AB}=4\sqrt{12}\cos(30)=12$$

good ! your answer is correct