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Vector Algebra

Question
CBSEENMA12034268
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Let vectors straight a with rightwards arrow on top space and space straight b with rightwards arrow on top be such that open vertical bar straight a with rightwards arrow on top close vertical bar space equals 3 space space space and space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space fraction numerator square root of 2 over denominator 3 end fraction comma then straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top is a unit vector, if the angle betweeen straight a with rightwards arrow on top space and space straight b with rightwards arrow on top is 

  • straight pi divided by 6
  • straight pi divided by 4
  • straight pi divided by 3
  • straight pi divided by 2

Solution

B.

straight pi divided by 4

Here,    open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 3 comma space space space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals fraction numerator square root of 2 over denominator 3 end fraction
Now,    straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top is a unit vector
i.e. if open vertical bar straight a with rightwards arrow on top space cross times space straight b with rightwards arrow on top close vertical bar space equals space 1
i.e. if open vertical bar straight a with rightwards arrow on top close vertical bar space open vertical bar straight b with rightwards arrow on top close vertical bar space space sin space straight theta space equals space 1 comma where straight theta is angle between straight a with rightwards arrow on top space and space straight b with rightwards arrow on top.
i.e. if  left parenthesis 3 right parenthesis space open parentheses fraction numerator square root of 2 over denominator 3 end fraction close parentheses space sin space straight theta space equals space 1
i.e.   if                     sin space straight theta space equals space fraction numerator 1 over denominator square root of 2 end fraction
i.e.   if                      straight theta space equals space straight pi over 4
therefore space space space space space left parenthesis straight B right parenthesis space is space correct space answer.

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