If $straight a with rightwards arrow on top space and space straight b with rightwards arrow on top$ are perpendicular, then $left parenthesis straight a with rightwards arrow on top plus straight b with rightwards arrow on top right parenthesis squared space equals space left parenthesis straight a with rightwards arrow on top minus straight b with rightwards arrow on top right parenthesis squared$

Solution

because space space straight a with rightwards arrow on top space and space straight b with rightwards arrow on top space space are space perpendicular because 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 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 1 right parenthesis

Now

open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses squared space equals space open parentheses straight a with rightwards arrow on top space minus space straight b with rightwards arrow on top close parentheses squared

straight a with rightwards arrow on top squared plus straight b with rightwards arrow on top squared space 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

i.e., if

2 space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space minus 2 space straight a with rightwards arrow on top. space straight b with rightwards arrow on top space space straight i. straight e. comma space space if space 4 straight a with rightwards arrow on top. space straight b with rightwards arrow on top

i.e., if

straight a with rightwards arrow on top. space straight b with rightwards arrow on top space equals space 0 comma

which is true.

open square brackets because space if space left parenthesis 1 right parenthesis close square brackets

Hence the result.

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