Let the position vectors of three vertices of a triangle be 4p+q-3r,-5p+q+2r and 2p-q+2r.

Question

Let the position vectors of three vertices of a triangle be $4\vec{p}+\vec{q}-3\vec{r},-5\vec{p}+\vec{q}+2\vec{r}$ and $2\vec{p}-\vec{q}+2\vec{r}$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p}+\vec{q}+\vec{r}}{4}$ and $\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}$ respectively, then $\alpha +2\beta +5\gamma $ is equal to :

Accepted Answer

We know that,

$\begin{gathered} O\frac{\overset{	o }{p}+\overset{	o }{q}+\overset{	o }{r}}{4} \ C(\alpha \overset{	o }{p}+\beta \overset{	o }{q}+\gamma \overset{	o }{r} \ C=\frac{\overset{	o }{p}+\overset{	o }{q}+\overset{	o }{r}}{3} \ \ \Rightarrow 2(\alpha \overset{	o }{p}+\beta \overset{	o }{q}+\gamma \overset{	o }{r})+\frac{\overset{	o }{p}+\overset{	o }{q}+\overset{	o }{r}}{4}=3(\frac{\overset{	o }{p}+\overset{	o }{q}+\overset{	o }{r}}{3}) \ \Rightarrow 8(\alpha \overset{	o }{p}+\beta \overset{	o }{q}+\gamma \overset{	o }{r})=3(\overset{	o }{p}+\overset{	o }{q}+\overset{	o }{r}) \ \Rightarrow 8\alpha =3,8\beta =3,8\gamma =3 \ \alpha =\frac{3}{8},\beta =\frac{3}{8},\gamma =\frac{3}{8} \ 	herefore \alpha +2\beta +3\gamma \ \frac{3}{8}+\frac{6}{8}+\frac{15}{8}=\frac{24}{8}=3 \end{gathered}$

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