Question:

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 a≠0,b≠0.

Updated On: Sep 19, 2023
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Solution and Explanation

(\(\vec a+\vec b\)).(\(\vec a+\vec b\))=\(\vec a\)|2+|\(\vec b\)|2
⇔ \(\vec a.\vec a\)+\(\vec a.\vec b\)+\(\vec b.\vec a\)+\(\vec b.\vec b\)=|\(\vec a\)|2+|\(\vec b\)|2 [Distributivity of scalar products over addition]
⇔|\(\vec a\)|2+2\(\vec a\).\(\vec b\)+|\(\vec b\)|2=|\(\vec a\)|2+|\(\vec b\)|2 [\(\vec a.\vec b=\vec b.\vec a\)(Scalar product is commutative)]
\(2\vec a.\vec b\)=0
\(\vec a.\vec b\)=0
\(\vec a\) and \(\vec b\) are perpendicular. [\(\vec a\)\(\vec 0\),\(\vec b\)\(\vec 0\)(Given)]

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