Let O be the origin and let PQR be an arbitrary triangle. The point S is such that.Then the triangle PQR has S as its

Engineering

Mathematics

Solving Vector Equations

Section Formulae and Centres of a Triangle

Question

Let O be the origin and let PQR be an arbitrary triangle. The point S is such that

$\vec{OP} \cdot \vec{OQ} + \vec{OR} \cdot \vec{OS} = \vec{OR} \cdot \vec{OP} + \vec{OQ} \cdot \vec{OS} = \vec{OQ} \cdot \vec{OR} + \vec{OP} \cdot \vec{OS}$ .

Then the triangle PQR has S as its

circumcentre

orthocenter

centroid

incentre

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Let p.v.s of P, Q, R & S be $\vec{p}, \vec{q}, \vec{r} & \vec{s}$

$∴ \vec{p} \cdot \vec{q} + \vec{r} \cdot \vec{s} = \vec{r} \cdot \vec{p} + \vec{q} \cdot \vec{s}$

$\Rightarrow \vec{p} \cdot (\vec{q} - \vec{r}) + \vec{s} \cdot (\vec{r} - \vec{q}) = 0 \Rightarrow (\vec{p} - \vec{s}) \cdot (\vec{q} - \vec{r}) = 0$

Similarly $(\vec{q} - \vec{s}) \cdot (\vec{r} - \vec{p})$ = 0

$∴$ Point S ( $\vec{s}$ ) is orthocentre of DPQR.