IIT JEE , AIEEE Forum - Mathematics , Vectors

bhuwanaroracorroded

let d

ABC. the points M,Nand P are taken on the sides AB, BC ,CA resp.

such that AM/AB=BN/BC=CP/CA=

suppose A as a origin

vectors AN,BP,CM are

a concurrent

b sides of a triangle

c sides of a isosceles triangle

d none of these

srujana

I think the question should be AN, BP , CM

Let A be the origin, the position vector of B be b, C be c and m,n for M and N.

given AM/BM =@

=> m-a = @(b-a)

m=@b .............. 1

lly for BN/BC

n= @c +b(1-@).....................2

for CP/CA

c(1-@) = p.............3

sub the values of @c and @b from 1 and 3 to 2

we get b+c = m+p+n

or as a=0 (origin)

a+b+c = m+p+n

(n-a) +(p-b) +(m-c) = 0

but n-a, p-b ,m-c are the position vectors of AN, BP, CM

thus AN+BP+CM=0

=> they are either the sides of a triangle or they are concurrent.

but clearly from the picture, they cant be the sides of a triangle, hence they are concurrent

bhuwanaroracorroded

hey but ur answer is wrong

d answer is b

i m also confused wid d answer

putting

=1/2 doesnt satisfy d answer

but if u take cross product of d sides in order to calculate d area

d area will not b equal to zero

bhuwanaroracorroded

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bhuwanaroracorroded

Re:vectors

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