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Question
If \[\vec{a}\] \[\vec{b}\] are two vectors such that \[\left| \vec{a} + \vec{b} \right| = \left| \vec{b} \right|\] then prove that \[\vec{a} + 2 \vec{b}\] is perpendicular to \[\vec{a}\]
Solution
\[\text{ Given that }\]
\[\left| \vec{a} + \vec{b} \right| = \left| \vec{b} \right|\]
\[\text{Squaring both sides, we get}\]
\[ \left| \vec{a} + \vec{b} \right|^2 = \left| \vec{b} \right|^2 \]
\[ \Rightarrow \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 + 2 \vec{a} . \vec{b} = \left| \vec{b} \right|^2 \]
\[ \Rightarrow \left| \vec{a} \right|^2 + 2 \vec{a} . \vec{b} = 0..............\left( 1 \right)\]
\[\text{ Now },\]
\[\left( \vec{a} + 2 \vec{b} \right) . \vec{a} \]
\[ = \vec{a} . \vec{a} + 2 \vec{b} . \vec{a} \]
\[ = \left| \vec{a} \right|^2 + 2 \vec{a} . \vec{b} \]
\[ = 0 .............\left[ \text{ Using } \left( 1 \right) \right]\]
\[\text{ So }, \vec{a} + 2 \vec{b} \text{ is perpendicular to } \vec{a} .\]
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