EM-3143

Vector geometry

E Math · Secondary 4 · Hard · Vector geometry

In triangle

ABC

, point

D

lies on

BC

such that

2BD = 3DC

.

Given

\vec{AB} = \mathbf{p}

and

\vec{AC} = \mathbf{q}

, express

\vec{AD}

in terms of

\mathbf{p}

and

\mathbf{q}

.

$\vec{AD} = \tfrac{2}{5}\mathbf{p} + \tfrac{3}{5}\mathbf{q}$
$\vec{AD} = \tfrac{3}{5}\mathbf{p} + \tfrac{2}{5}\mathbf{q}$
$\vec{AD} = \tfrac{3}{5}\mathbf{q} - \tfrac{2}{5}\mathbf{p}$
$\vec{AD} = \tfrac{2}{5}\mathbf{q} - \tfrac{3}{5}\mathbf{p}$

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