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

Question

Aksh · Answer

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 · Answer

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 = &ndash; a / b &hellip;.1 As, resultant(r) = (a 2 + b 2 + 2ab cos z) 1/2 so, by putting the value of cos z in this equation from equation 1, we get resultant(r) = ( b 2 - a 2 ) 1/2 NOW A.T.Q., WE PUT THE RESULTANT VALUE IN THIS EQUATION, b / 2 = (b 2 &ndash; a 2 ) 1/2 squaring both sides, we get (b 2 / 4) &ndash; b 2 = &ndash; a 2 therefore, a = (root3 / 2) * b so, a/b = root3 / 2 PUT THIS IN EQUATION 1, WE GET cos z = &ndash; ROOT3 / 2 THEREFORE, Z = 150 0

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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

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

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