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Mechanical Properties Of Fluids

Question
CBSEENPH11020661
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If space vectors space straight A space equals space space cos space ωt space straight i with hat on top space plus space sin space straight omega straight j with hat on top space and space straight B space equals space cos space ωt over 2 straight i with hat on top space plus space sin ωt over 2 straight j with hat on top are functions of time, then the value of t at which they are orthogonal to each other
  • straight t space equals space fraction numerator straight pi over denominator 4 straight omega end fraction
  • straight t space equals space fraction numerator straight pi over denominator 2 straight omega end fraction
  • t equals space straight pi over straight omega

  • t = 0 

Solution

C.

t equals space straight pi over straight omega

For a perpendicular vector, we have A. B  = 0 
left square bracket cos space ωt space straight i with hat on top space plus space sin space ωt space straight j with hat on top right square bracket space open square brackets cos space ωt over 2 straight i with hat on top space plus space fraction numerator sin space ωt over denominator 2 end fraction space straight j with hat on top close square brackets space equals space 0 rightwards double arrow space left square bracket therefore space cos space left parenthesis straight A minus straight B right parenthesis space equals space cos space straight A space Cos space straight B space plus space sin space straight A space Sin space straight B right square bracket cos space open parentheses ωt space minus space ωt over 2 close parentheses space equals space 0 cos space ωt over 2 space equals space 0 space rightwards double arrow space ωt over 2 space equals space straight pi over 2 space rightwards double arrow space straight t space equals space straight pi over straight omega Thus comma space time space taken space by space vectors space which space are space orthogonal space to space each space other space is space straight pi over straight omega

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