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Question
Using vector method, find incentre of the triangle whoose vertices are P(0, 4, 0), Q(0, 0, 3)
and R(0, 4, 3).
Solution
Let `barp, barq ,barr` be the position vectors of vertices P, Q, R of Δ PQR respectively
`barp=4hatj,barq=3hatk,barr=4hatj+3hatk`
`bar(PQ)=barq-barp=3hatk-4hatj=-4hatj+3hatk`
`bar(QR)=barr-barq=4hatj+3hatk-3hatk=4hatj`
`bar(RP)=barp-barr=4hatj-4hatj-3hatk=-3hatk`
Let x, y, z be the lengths of opposites of vertices P,Q,R respectively.
`x=|bar(QR)|=4`
`y=|bar(RP)|=3`
`z=|bar(PQ)|=sqrt(16+9)=sqrt25=5`
If H(`barh`)is the incentre of `Delta`PQR then
`barh=(xbarp+ybarq+zbarr)/(x+y+z)`
`=(4(4hatj)+3(3hatk)+5(4hatj+3hatk))/(4+3+5)`
`=(16hatj+9hatk+20hatj+15hatk)/12`
`=(36hatj+24hatk)/12=3hatj+2hatk`
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