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Vector Algebra

Question
CBSEENMA12034087
Wired Faculty App

Let straight a with rightwards arrow on top space and space straight b with rightwards arrow on top be the two non-zero vectors. Then, prove that straight a with rightwards arrow on top space and space straight b with rightwards arrow on top are perpendicular if and only 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 open vertical bar straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close vertical bar.

Solution

(i) Assume that straight a with rightwards arrow on top space and space straight b with rightwards arrow on top are perpendicular
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 0                                                    ...(i)
             open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar squared space equals space left parenthesis straight a with rightwards arrow on top plus 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 plus 2 space straight a with rightwards arrow on top. space straight b with rightwards arrow on top
equals space straight a with rightwards arrow on top squared plus straight b with rightwards arrow on top squared space plus space space 2 space left parenthesis 0 right parenthesis space space space space space space space space space space space space space space open square brackets because space space of space left parenthesis straight i right parenthesis close square brackets
therefore space space space space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar squared space equals space straight a with rightwards arrow on top squared plus straight b with rightwards arrow on top squared space                                       ...(2)
Again,  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 minus 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 straight a with rightwards arrow on top. space straight b with rightwards arrow on top
                             equals space straight a with rightwards arrow on top squared plus straight b with rightwards arrow on top squared space minus space 2 left parenthesis 0 right parenthesis space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis straight i right parenthesis close square brackets
therefore space space space 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 straight a with rightwards arrow on top squared plus straight b with rightwards arrow on top squared                                         ...(3)
Fom (2) and (3), we get.
         open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar squared space equals space open vertical bar straight a with rightwards arrow on top space minus straight b with rightwards arrow on top close vertical bar squared
therefore space space space space space open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar space equals space open vertical bar straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close vertical bar
(ii) Assume that 
              open vertical bar straight a with rightwards arrow on top plus straight b with rightwards arrow on top close vertical bar space equals space open vertical bar straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close vertical bar
therefore space space space space open vertical bar straight a with rightwards arrow on top space plus space straight b with rightwards arrow on top close vertical bar squared space equals space open vertical bar straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close vertical bar squared                                       open square brackets because space space straight a with rightwards arrow on top squared space equals space open vertical bar straight a with rightwards arrow on top close vertical bar squared close square brackets
rightwards double arrow space space space space space straight a with rightwards arrow on top squared plus straight b with rightwards arrow on top squared plus 2 space straight a with rightwards arrow on top. space straight b with rightwards arrow on top 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 rightwards double arrow space space space space space space 2 space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space space equals space minus space 2 space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space space space space rightwards double arrow space space space 4 space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0 rightwards double arrow space space space space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0 space space space space rightwards double arrow space space space space straight a with rightwards arrow on top space and space straight b with rightwards arrow on top space are space perpendicular.

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