A straight line `L` cuts the lines `A B ,A Ca n dA D` of a parallelogram `A B C D` at points `B_1, C_1a n dD_1,` respectively. If `( vec A B)_1,lambda_1 vec A B ,( vec A D)_1=lambda_2 vec A Da n d( vec A C)_1=lambda_3 vec A C ,` then prove that `1/(lambda_3)=1/(lambda_1)+1/(lambda_2)` .

A straight line L cuts the sides AB, AC, AD of a parallelogram ABCD at B_(1), C_(1), d_(1) respectively. If vec(AB_(1))=lambda_(1)vec(AB), vec(AD_(1))=lambda_(2)vec(AD) and vec(AC_(1))=lambda_(3)vec(AC), then (1)/(lambda_(3)) equal to

A B C D parallelogram, and A_1a n dB_1 are the midpoints of sides B Ca n dC D , respectivley . If vec AA_1+ vec A B_1=lambda vec A C ,t h e nlambda is equal to a. 1/2 b. 1 c. 3/2 d. 2 e. 2/3

A B C D isa parallelogram with A C a n d B D as diagonals. Then, vec A C- vec B D= 4 vec A B b. 3 vec A B c. 2 vec A B d. vec A B

If [vec a xxvec bvec d xxvec cvec c xxvec a] = lambda [vec with bvec c] ^ (2) then lambda is equal to

If vec a and vec b are two vectors and lambda>0, then the least value of (1+lambda)|vec a|^(2)+(1+(1)/(lambda))|vec b|^(2) is

If D,E and F are the mid-points of the sides BC, CA and AB respectively of a triangle ABC and lambda is scalar, such that vec(AD) + 2/3vec(BE)+1/3vec(CF)=lambdavec(AC) , then lambda is equal to

If A,B and are collinear find lambda non-collinear such that vec c=(lambda-2)vec a+vec b and vec d=(2 lambda+1)vec a-vec b,vec c and vec d are collinear find lambda

Write the equation of the plane containing the lines vec r= vec a+lambda vec b\ a n d\ vec r= vec a+mu vec c

vec r = lambda (vec a xxvec b) + mu (vec b xxvec c) + gamma (vec c xxvec a) and [vec with bvec c] = (1) / (8), then the value of lambda + mu + gamma =

The position vectors of the vertices A ,Ba n dC of a triangle are three unit vectors vec a , vec b ,a n d vec c , respectively. A vector vec d is such that vecd dot vec a= vecd dot vec b= vec d dot vec ca n d vec d=lambda( vec b+ vec c)dot Then triangle A B C is a. acute angled b. obtuse angled c. right angled d. none of these