Vectors and triangles

Question

ok so the question says: given triangle $ABC$ and a point $D$ defined by: $$\vec{AD}= 3 \vec{AC} + \tfrac{1}{2}\vec{CB}$$ Draw the figure and prove that vector $$\vec{BD} = -2\vec{BA} + \tfrac{5}{2}\vec{BC}$$

Answer 1

Hint: Use that $\vec{AX}=\vec{AB}+\vec{BX}$ for any point $X$.

With this you can fix one of these points, say $A$ and start every vector from there. FDor example you can write $\vec{CB}=\vec{CA}+\vec{AB} = -\vec{AC}+\vec{AB}$

Answer 2

We have

$ \begin{align*} \overrightarrow{BD} & = \overrightarrow{BA}+\overrightarrow{AD} \\ & = \overrightarrow{BA}+3 \overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CB} \\ & = \overrightarrow{BA}+3 \left( \overrightarrow{BC}-\overrightarrow{BA} \right)- \dfrac{1}{2}\overrightarrow{BC} \\ & = -2\overrightarrow{BA}+\dfrac{5}{2}\overrightarrow{BC}. \end{align*} $