Relate external division, area ratios and a unit-vector projection

$OB=(2OA+3OC)/5$. Hence $OC=(5OB-2OA)/3=\frac53\mathbf b-\frac23\mathbf a$.

$\operatorname{area}(OBC)=\frac12|\mathbf b\times(\frac53\mathbf b-\frac23\mathbf a)|=\frac12|\mathbf b\times(-\frac23\mathbf a)|=\frac13|\mathbf a\times\mathbf b|$, using $\mathbf b\times\mathbf b=0$ and reversal of cross-product order. Thus k=1/3.

Triangles OBC and OAB share the height to line ABC. Since BC/AB=2/3, area (OBC)=$\frac23\operatorname{area}(OAB)=\frac23(\frac12|\mathbf a\times\mathbf b|)=\frac13|\mathbf a\times\mathbf b|$.

Write $OD=\lambda\mathbf b$ with 0<λ<1. Then area (OAD)=$\frac12|\mathbf a\times\lambda\mathbf b|=\frac\lambda2|\mathbf a\times\mathbf b|$. Dividing by area (OBC) gives $(\lambda/2)/(1/3)=4/3$, hence λ=8/9 and $OD=\frac89\mathbf b$. The supplied area ratio presupposes nondegenerate triangles.

Because a is unit, the projection length is $|OC\cdot\mathbf a|=|\frac53\mathbf b\cdot\mathbf a-\frac23\mathbf a\cdot\mathbf a|=|\frac53|\mathbf b||\mathbf a|\cos60^\circ-\frac23|\mathbf a|^2|=|5/6-2/3|=1/6$.

$OA=OB=1$ and angle AOB=60° make OAB equilateral, so AB=1 and $AC=\frac53AB=5/3$. The projection length of AC onto OA is $AC\cos\angle OAB=(5/3)\cos60^\circ=5/6$. The perpendicular foot of C is 1−5/6=1/6 from O along OA, giving the same projection length.