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Vector Algebra

Question
CBSEENMA12034161
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Prove, using vectors, that the altitudes of a triangle are concurrent.
OR
Prove that the perpendicular from the vertices to the opposite sides of a triangle are concurrent. 

Solution

Let ABC be given triangle and H be the point of intersection of altitudes AL and BM. Join CH and produce it to meet BA in N.
Take H as origin.

   Let straight a with rightwards arrow on top comma space straight b with rightwards arrow on top comma space stack straight c with rightwards arrow on top with rightwards arrow on top be the position vectors of A, B, C respectively.
therefore space space space space space HA with rightwards arrow on top space equals space straight a with rightwards arrow on top comma space space HB with rightwards arrow on top space equals space straight b with rightwards arrow on top comma space space HC with rightwards arrow on top space equals space straight c with rightwards arrow on top.
Since AL space perpendicular space BC
  therefore space space space space AH with rightwards arrow on top. space space BC with rightwards arrow on top space equals space 0 therefore space space space space minus straight a with rightwards arrow on top. space space left parenthesis straight c with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis space equals space 0 therefore space space space space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top. space space straight c with rightwards arrow on top space space equals space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Again,  BM perpendicular CA
therefore space space space space BH with rightwards arrow on top space. space CA with rightwards arrow on top space equals space 0 rightwards double arrow space space minus space stack straight b. with rightwards arrow on top space left parenthesis straight a with rightwards arrow on top space minus space straight c with rightwards arrow on top right parenthesis space equals space space 0 space space space space space space rightwards double arrow space space space space straight b with rightwards arrow on top. space straight c with rightwards arrow on top space minus space straight b with rightwards arrow on top. space straight a with rightwards arrow on top space equals space 0
rightwards double arrow space space space space space straight b with rightwards arrow on top. space straight c with rightwards arrow on top space minus space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0                                                   ...(2)
Adding (1) and (2), we get,
            straight b with rightwards arrow on top. space straight c with rightwards arrow on top space minus space straight a with rightwards arrow on top. space straight c with rightwards arrow on top space equals space space 0 space space space space space space space space rightwards double arrow space space space space left parenthesis straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top right parenthesis. space straight c with rightwards arrow on top space equals space 0
rightwards double arrow space space minus space straight c with rightwards arrow on top. space thin space left parenthesis straight a with rightwards arrow on top space minus space stack straight b right parenthesis with rightwards arrow on top space equals space 0 space space space space rightwards double arrow space space space space CH with rightwards arrow on top. space space BA with rightwards arrow on top space equals space 0 rightwards double arrow space space space space space space CH space perpendicular space AB space space space space space space space space space space space space space space space space space rightwards double arrow space space space space CN thin space perpendicular space AB therefore space space space space space space space space space space altitudes space of space increment ABC space meet space in space straight a space point space straight H.

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