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Q15  Prove that ( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a ^2 | + |\vec b |^2 , if and only if  \vec a , \vec b are perpendicular, given \vec a \neq 0 , \vec b \neq 0
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Given in the question,

\vec a , \vec b are perpendicular and we need to prove that ( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a ^2 | + |\vec b |^2

LHS= ( \vec a + \vec b ) . (\vec a + \vec b ) = \vec a .\vec a+\vec a.\vec b+\vec b.\vec a+\vec b.\vec b

                                = \vec a .\vec a+2\vec a.\vec b+\+\vec b.\vec b

                                = |\vec a |^2+2\vec a.\vec b+\+|\vec b|^2

if  \vec a , \vec b are perpendicular,  \vec a.\vec b=0

 ( \vec a + \vec b ) . (\vec a + \vec b ) = |\vec a |^2+2\vec a.\vec b+\+|\vec b|^2

                                 = |\vec a |^2+2\cdot0+\+|\vec b|^2

                                 = |\vec a |^2+|\vec b|^2

                                 = RHS

 LHS ie equal to RHS

Hence proved.

Posted by

Pankaj Sanodiya

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