`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`

G is the centroid of triangle A B Ca n dA_1a n dB_1 are rthe midpoints of sides A Ba n dA C , respectively. If "Delta"_1 is the area of quadrilateral G A_1A B_1a n d"Delta" is the area of triangle A B C , then "Delta"//"Delta"_1 is equal to a.3/2 b. 3 c. 1/3 d. none of these

If a :(b+c)=1:3 a n d c :(a+b)=5:7, t h e b :(a+c) is equal to 1:2 b. 1:3 c. 2:3 d. 2:1

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

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) .

ABCDEF si a regular hexagon with centre at the origin such that A vec D + E vec B + F vec C = lambda E vec D . " Then, " lambda equals

If A : B=1/2:1/3n a d B : C=1/2:1/3, t h e n A : B : C is equal to 1:2:6 b. 3:2:6 c. 2:3:3 d. 9:6:4

Let vec a,vec b and vec c be three vectors having magnitudes 1,5 and 3, respectively,such that the angel between vec a and vec b is theta and vec a xx(vec a xxvec b)=c. Then tan theta is equal to a 0 b.2/3 c.3/5 d.3/4

Vectors vec a\ a n d\ vec b are inclined at angel theta=120^0 . If | vec a|=1,\ | vec b|=2,\ then [( vec a+3 vec b)xx(3 vec a- vec b)]^2 is equal to 300 b. 235 c. 275 d. 225

A B C D E is pentagon, prove that vec A B + vec B C + vec C D + vec D E+ vec E A = vec0 vec A B+ vec A E+ vec B C+ vec D C+ vec E D+ vec A C=3 vec A C

If | vec axx vec b|=4,\ | vec adot vec b|=2,\ t h e n\ | vec a|^2| vec b|^2= 6 b. 2 c. 20 d. 8