IIT JEE , AIEEE Forum - Mathematics , Vectors

karthik_abiram

if in a rt triangle ABC ,the hypotenuse AB=p ,then then

AB.AC+BC.BA+CA.CB=

[all in vectors]

1]2p²

2]p²

3]p²/2

4]2p

i get it as p² ,but it is given as 2p²

please help

ankurgupta91

let CA=r ,CB=q(vector)
and # is angle CBA
nw AB.AC+BC.BA+CA.CB
=prsin#+pqcos#+qrcos90
p=rsin# , q=rcos#
=r^2+q^2
=p^2
so ur ans is correct
thanks.......

vasanth

AB.AC + BC.BA + CA.CB = p.(psin$).cos(90 - $) + p.(pcos$)cos$

= p²( sin²$ + cos²$)

=p²

so there is no way u can get 2p²........u r absolutely correct karthik

cheeeeeeeers

neeraj_agarwal_1990

The ans is p^2....100000%

karthik_abiram

thank you every one

rinkupark

p^2 is correct

vasanth

IABI = p

IACI = p sin$

IBCI = p cos$

IBAI = p

AB.AC + BC.BA + CB.CA = p IACI .cos (90 - $) + p IBCI cos$ +ICAIICBIcos90

= p. psin$ sin$ + p. pcos$ cos$ + 0

= p² (sin²$ + cos²$)

= p²

hope u got it......

cheeeeeers

karthik_abiram

thanx 4 attaching the picture vasanth

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