The discussion focuses on calculating the dot product of vectors in rectangle $ABCD$, specifically $\vec{AC} \cdot \vec{AB}$. Given that $\angle CAB = 30^\circ$ and the condition $\vec{AC} \cdot \vec{AD} = |\vec{AC}|$, the relationship between the vectors is established. The conclusion confirms that the calculation of $\vec{AC} \cdot \vec{AB}$ is accurate and aligns with the geometric properties of the rectangle.
PREREQUISITESStudents of geometry, mathematicians, and anyone interested in vector analysis within geometric shapes.
good ! your answer is correctgreg1313 said:$$\vec{AC}\cdot\vec{AD}=|\vec{AC}||\vec{AD}|\cos(60)=\frac12|\vec{AC}||\vec{AD}|=|\vec{AC}|\implies|\vec{AD}|=2$$
$$|\vec{AC}|\cos(60)=2\implies|\vec{AC}|=4\implies|\vec{AB}|=\sqrt{12}$$
$$\vec{AC}\cdot\vec{AB}=4\sqrt{12}\cos(30)=12$$