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Question
Find \[\left| \vec{a} \right| and \left| \vec{b} \right|\] if
\[\left( \vec{a} + \vec{b} \right) \cdot \left( \vec{a} - \vec{b} \right) = 3\text{ and } \left| \vec{a} \right| = 2\left| \vec{b} \right|\]
Solution
\[\ \text{ Given that }\]
\[\left| \vec{a} \right| = 2 \left| \vec{b} \right| . . . \left( 1 \right)\]
\[\text{ And } \left( \vec{a} + \vec{b} \right) . \left( \vec{a} - \vec{b} \right) = 3\]
\[ \Rightarrow \left| \vec{a} \right|^2 - \left| \vec{b} \right|^2 = 3\]
\[ \Rightarrow \left( 2 \left| \vec{b} \right| \right)^2 - \left| \vec{b} \right|^2 = 3........... \left[ \text{ From } (1) \right]\]
\[ \Rightarrow 4 \left| \vec{b} \right|^2 - \left| \vec{b} \right|^2 = 3\]
\[ \Rightarrow 3 \left| \vec{b} \right|^2 = 3\]
\[ \Rightarrow \left| \vec{b} \right|^2 = 1\]
\[ \Rightarrow \left| \vec{b} \right| = 1\]
\[\left| \vec{a} \right| = 2 \left| \vec{b} \right| = 2 \left( 1 \right) = 2\]
\[ \therefore \left| \vec{a} \right| = 2 \text{ and } \left| \vec{b} \right| = 1\]
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