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