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Question
Using vectors prove that the altitudes of a triangle are concurrent.
Solution
Let, the altitudes AD and BE intersect at O
Join CO and produce to meet AB in F
Let `vec(OA) = veca`
`vec(OB) = vecb. vec(OC) = vecc`
`veca` is perpendicular to `vec(BC) = vec(OC) = vecc - vecb`
Therefore, `veca. (vecc - vecb)` = 0
⇒ `veca.vecb = veca.vecb` ......(i)
Also direction OB is perpendicular to AC.
∴ `vecb.(vecc - veca)` = 0
⇒ `vecb. vecc = veca . vecb` ......(ii)
From equations (i) and (ii), we get
`veca.vecc = veca.vecb = vecb.vecc`
i.e., `vecc.veca - vecc.vecb` = 0
⇒ `vecc. (veca - vecb)` = 0 ......(iii)
`veca - vecb = vec(OA) - vec(OB) = vec(BA)`
And by virtue of equation (iii), `vecC` is perpendicular to ``vec(BA), but `vecC` is a vector in direction of `vec(OC)`
Hence, `vec(OC)` is perpendicular to `vec(AB)`, i.e., CF is the third altitude of the triangle through C.
Hence. the 3 altitudes are concurrent at O.
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