# If A-B perpendicular to B and |A-B|=A/2 angle between A,B is

Akash Kumar Dutta
98 Points
11 years ago

B

/

/ x degree

/___________  A

/ 90\

/         \  A/2

/ x           \

/____________\

-B        A         A-B

let the angle between A and B vector be x.

then we know A-B is perpendicular to B so its also perpendicular to -B.

hence  a triangle forms by -B , A-B , and A.in which angle between -B and A is x

and angle between A-B and -B is 90.

applying sine rule.

|A|/sin 90= |A-B|/sin x

solving we get x=30 degree (ANS).

hope i helped you.

gauhar singh
7 Points
11 years ago

|A-B|=A/2 i.e. A2+B2 - 2ABCos@ = A2/4------------>(1)

also, (A-B).B = 0 as angle bwtween is 90 (dot product |a.b|=|a||b|Cos@ )

ABCos@ = B2---------->(2)

on substituting in (1) from (2) we get A = 2/(3)1/2B

again on substituing in (2) we get Cos@= (3)1/2/2

hence @ = 30

lokesh soni
37 Points
11 years ago

A-B PERPENDICULAR TO B

LET ANGLE BETWEEN A AND B BE θ

THEN BSinθ/(A-BCosθ)=tan 90degree=1/0

that gives A-BCosθ=0

A=BCosθ

A/B=COSθ

and also

√A2+B2-2ABCOSθ=lA-Bl=A/2

√A2+B2-2AA=A/2                  (BCOSθ=A)

B2-A2=A2/4

4B2=5A2

A2/B2=COS2 θ=4/5

COSθ=2/√5