Home
Class 12
MATHS
A straight line L cuts the lines A B ,A ...

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 1/(lambda_3) .

A

`(1)/(lambda_1)+(1)/(lambda_2)`

B

`(1)/(lambda_1)-(1)/(lambda_2)`

C

`-(lambda_1)+(lambda_2)`

D

`(lambda_1)+(lambda_2)`

Text Solution

Verified by Experts

The correct Answer is:
(a)
Promotional Banner

Topper's Solved these Questions

  • THREE DIMENSIONAL COORDINATE SYSTEM

    ARIHANT MATHS|Exercise JEE Type Solved Examples : Matching Type Questions|5 Videos
  • THREE DIMENSIONAL COORDINATE SYSTEM

    ARIHANT MATHS|Exercise Exercise For Session 1|12 Videos
  • THEORY OF EQUATIONS

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|35 Videos
  • TRIGONOMETRIC EQUATIONS AND INEQUATIONS

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|12 Videos

Similar Questions

Explore conceptually related problems

If a, b and c are non-coplanar vectors and d=lambda vec a+mu vec b+nu vec c , then lambda is equal to

Vectors drawn the origin O to the points A , B and C are respectively vec a , vec b and vec4a- vec3bdot find vec A C and vec B Cdot

Prove that [vec a -vec b, vec b - vec c, vec c- vec a] =0

Prove that [vec a+ vec b, vec b +vec c, vec c + vec a]= 2 [vec a vec b vec c]

In a regular hexagon A B C D E F ,\ A vec B=a ,\ B vec C= vec b\ a n d\ vec C D=c\ T h e n\ vec A E=

If vec c is normal to vec a and vec b , show that vec c is normal to vec a+ vec b and vec a - vec b .

Statement 1: Let vec a , vec b , vec c and vec d be the position vectors of four points A ,B ,Ca n dD and 3 vec a-2 vec b+5 vec c-6 vec d=0. Then points A ,B ,C ,a n dD are coplanar. Statement 2: Three non-zero, linearly dependent coinitial vector ( vec P Q , vec P Ra n d vec P S) are coplanar. Then vec P Q=lambda vec P R+mu vec P S ,w h e r elambdaa n dmu are scalars.

Statement 1: if three points P ,Qa n dR have position vectors vec a , vec b ,a n d vec c , respectively, and 2 vec a+3 vec b-5 vec c=0, then the points P ,Q ,a n dR must be collinear. Statement 2: If for three points A ,B ,a n dC , vec A B=lambda vec A C , then points A ,B ,a n dC must be collinear.

In a trapezium ABCD the vector B vec C = lambda vec(AD). If vec p = A vec C + vec(BD) is coillinear with vec(AD) such that vec p = mu vec (AD), then

If | vec a|=2,\ | vec b|=5\ a n d\ | vec axx vec b|=8 , find vec adot vec bdot