# The resultant of the vectors A and B is perpendicular to the vector A and it`s magnitude is equal to half the magnitude of vector B the angle between A and B is

Grade:11

## 2 Answers

Aksh
31 Points
7 years ago
Let the angle between the two vectors A & B be x And angle between A vector And resultant vector be yTherefore ,tany=Bsinx/A+BcosxBut y=90 giventan90=Bsinx/A+BcosxBut tan90= infinity =1/01/0=Bsinx/A+BcosxCross multiply A+Bcosx=0cosx=-A/BBut |A+B|=B/2 given √A×A+B×B+2ABcosx=B/2Thus,√A×A+B×B-2AB(-A/B)=B/2√A×A+B×B-2A×A=B/2Squaring both sides B×B-A×A=B×B/2A×A=3/4B×BTaking square rootA=B√3/2Cosx = -A/B=-B√3/2÷B=-√3/2Cosx =-√3/2Therefore, x=150Thus angle between A&B vectors is 150
ANIRUDH VATS
13 Points
6 years ago
SOLUTION: As, the angle between the resultant and vector A is 90.(x)
THEREFORE,
tan x = b sin z / a + b cos z
so,     tan90 = b sin z / a + b cos z
1/0     = b sin z / a + b cos z
by reciprocal
0/1    = a + b cos z /  b sin z
so,       a + b cos z = 0
cos z = – a / b                 ….1
As, resultant(r) = (a2 + b2 + 2ab cos z) 1/2
so, by putting the value of cos z  in this equation from equation 1, we get
resultant(r) = ( b- a2)1/2
NOW A.T.Q., WE PUT THE RESULTANT VALUE IN THIS EQUATION,
b / 2 = (b2 – a2)1/2
squaring both sides, we get
(b/ 4) – b= – a2
therefore,
a = (root3 / 2) * b
so,
a/b = root3 / 2
PUT THIS IN EQUATION 1, WE GET
cos z = – ROOT3 / 2
THEREFORE,
Z = 1500

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