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Vector Algebra

Question
CBSEENMA12034094
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If straight a with rightwards arrow on top space and space straight b with rightwards arrow on top are unit vectors, and straight theta is the angle between them, find 1 half open vertical bar straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close vertical bar in terms of straight theta.

Solution

We have
      open vertical bar straight a with rightwards arrow on top minus straight b with rightwards arrow on top close vertical bar squared space equals space left parenthesis straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis squared space equals space straight a with rightwards arrow on top squared plus straight b with rightwards arrow on top squared space minus space 2 space straight a with rightwards arrow on top. space straight b with rightwards arrow on top
                     equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared plus space open vertical bar straight b with rightwards arrow on top close vertical bar squared space minus space 2 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 equals space left parenthesis 1 right parenthesis squared plus left parenthesis 1 right parenthesis squared space minus space 2 left parenthesis 1 right parenthesis thin space left parenthesis 1 right parenthesis space cos space straight theta
                                            open square brackets because space open vertical bar straight a with rightwards arrow on top close vertical bar space equals space 1 comma space space open vertical bar straight b with rightwards arrow on top close vertical bar space equals space 1 space as space straight a with rightwards arrow on top comma space straight b with rightwards arrow on top space are space unit space vectors close square brackets
                 equals space 2 left parenthesis 1 minus cos space straight theta right parenthesis space equals space 2 space open parentheses 2 space sin squared space straight theta over 2 close parentheses
therefore space space space space open vertical bar straight a with rightwards arrow on top space minus straight b with rightwards arrow on top close vertical bar squared space equals space 4 space sin squared space straight theta over 2 therefore space space space space sin space straight theta over 2 space equals 1 half space open vertical bar straight a with rightwards arrow on top minus straight b with rightwards arrow on top close vertical bar space space space space space space space space space space space open square brackets because space space 0 space less or equal than space straight theta space less or equal than space straight pi space space rightwards double arrow space space sin space straight theta over 2 greater or equal than space 0 close square brackets because space space space space 1 half open vertical bar straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close vertical bar space equals space sin space straight theta over 2.
    

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