Home
Class 12
MATHS
If two vectors OA and OB are there, then...

If two vectors OA and OB are there, then their resultant OA+OB can be found by completin the parallelogram OACB and OC=OA+OB. Also, if |OA|=|OB|, then the resultant will bisect the angle between them.
Q. If internal and external bisectors of `angleA` of `DeltaABC` meet the base BC at D and E respetively, then (D and E lie on samme side of B).

A

`BC=(BD+BE)/(4)`

B

`BC^(2)=BDxxDE`

C

`(2)/(BC)=(1)/(BD)+(1)/(BE)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
C
Here, `(AB)/(AC)=(BD)/(DC) and (AB)/(AC)=(BE)/(CE)`

`implies (BD)/(DC)=(BE)/(CE) implies (BD)/(CE)implies (BD)/(BC-BD)=(BE)/(BE-BC)`
`implies BD*BE-BD*BC=BC*BE-BD*BE`
`implies 2BD*BE=(BD+BE)*BC`
or `(2)/(BC)=(1)/(BD)+(1)/(BE)`.
Promotional Banner

Topper's Solved these Questions

  • VECTOR ALGEBRA

    ARIHANT MATHS|Exercise Exercise (Matching Type Questions)|1 Videos

  • VECTOR ALGEBRA

    ARIHANT MATHS|Exercise Exercise (Single Integer Answer Type Questions)|7 Videos

  • VECTOR ALGEBRA

    ARIHANT MATHS|Exercise Exercise (Statement I And Ii Type Questions)|3 Videos

  • TRIGONOMETRIC FUNCTIONS AND IDENTITIES

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|19 Videos

Similar Questions

Explore conceptually related problems

If two vectors OA and OB are there, then their resultant OA+OB can be found by completin the parallelogram OACB and OC=OA+OB. Also, if |OA|=|OB|, then the resultant will bisect the angle between them. Q. A vector C directed along internal bisector of angle between vectors A=7hati-4hatj-4hatk and B=-2hati-hatj+2hatk with |C|=5sqrt(6) is

if D,E and F are the mid-points of the sides BC,CA and AB respectively of the DeltaABC and O be any points, then prove that OA+OB+OC=OD+OE+OF

Find the resultant of the three vectors vec OA,vec OB and vec OC each of magnitude r as shown in figure?

Two adjacent sides OA and OB of a rectangle OACB are represented by vec a and vecb respectively, where o is origin. If 16|vec axxvecb| = 3 |veca+vecb|^2 and theta is the angle between the diagonals OC and AB, then the value(s) of tan(theta/2)

A (1, -2) and B (2, 5) are two points. The line OA, OB are produced to C and D respectively such that OC=2OA and OD=20B . Find CD