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Vector Algebra

Question
CBSEENMA12034141
Wired Faculty App

Let straight a with rightwards arrow on top space and space straight b with rightwards arrow on top be two unit vectors and straight theta be the angle between them. Then straight a with rightwards arrow on top plus straight b with rightwards arrow on top is a unit vector if choose the correct answer.

  • straight theta space equals straight pi over 4
  • straight theta space equals space straight pi over 3
  • straight theta space equals space straight pi over 2
  • straight theta space equals fraction numerator 2 straight pi over denominator 3 end fraction

Solution

D.

straight theta space equals fraction numerator 2 straight pi over denominator 3 end fraction

Since straight a with rightwards arrow on top space and space straight b with rightwards arrow on top are unit vectors.
therefore space space space space open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 1 comma space space space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals 1                                  
Also,  straight theta is angle between straight a with rightwards arrow on top space and space straight b with rightwards arrow on top.
therefore space space space space space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 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 cos space straight theta space equals space left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis space cos space straight theta or space space space space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space cos space straight theta space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis Now space straight a with rightwards arrow on top plus straight b with rightwards arrow on top space is space straight a space unit space vector
if    open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar space equals space 1
i.e. if  open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses. space left parenthesis straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top right parenthesis space equals space 1
i.e. if straight a with rightwards arrow on top. space straight a with rightwards arrow on top space plus space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space plus straight b with rightwards arrow on top. space straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top. straight b with rightwards arrow on top space equals 1
i.e. if open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space plus space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space plus space open vertical bar straight b with rightwards arrow on top close vertical bar squared space equals space 1
i.e if left parenthesis 1 right parenthesis squared plus space cos space straight theta space plus space cos space straight theta space plus space 1 space equals space 1 space space space space space space space space space space space space space space space open square brackets because space space of space left parenthesis 1 right parenthesis comma space space left parenthesis 2 right parenthesis close square brackets
i.e. if     2 space cos space straight theta space equals space minus 1
i.e.  if     cos space straight theta space equals space minus 1 half
i.e.  if    straight theta space equals space fraction numerator 2 space straight pi over denominator 3 end fraction                            open square brackets because space minus 1 half space equals negative cos space 60 degree space equals space cos space left parenthesis 180 degree minus 60 degree right parenthesis space equals space cos space 120 degree space equals space cos space fraction numerator 2 space straight pi over denominator 3 end fraction close square brackets
therefore space space space left parenthesis straight D right parenthesis space is space correct space answer.

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