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Question
If \[\vec{a} \text{ and } \vec{b}\] are two non-collinear unit vectors such that \[\left| \vec{a} + \vec{b} \right| = \sqrt{3},\] find \[\left( 2 \vec{a} - 5 \vec{b} \right) \cdot \left( 3 \vec{a} + \vec{b} \right) .\]
Solution
\[\text{ We } have\]
\[\left| \vec{a} + \vec{b} \right| = \sqrt{3}\]
\[\text{ Squaring both sides , we get } \]
\[ \left| \vec{a} + \vec{b} \right|^2 = 3\]
\[ \Rightarrow \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 + 2 \vec{a} . \vec{b} = 3\]
\[ \Rightarrow 1 + 1 + 2 \vec{a} . \vec{b} = 3 ...........(\text{ Because } \vec{a} \text{ and } \vec{b} \text{ are unit vectors) }\]
\[ \Rightarrow 2 + 2 \vec{a} . \vec{b} = 3\]
\[ \Rightarrow 2 \vec{a} . \vec{b} = 1\]
\[ \Rightarrow 2 \vec{a} . \vec{b} = 1\]
\[ \Rightarrow \vec{a} . \vec{b} = \frac{1}{2} . . . \left( 1 \right)\]
\[\text{ Now },\]
\[\left( 2 \vec{a} - 5 \vec{b} \right) . \left( 3 \vec{a} + \vec{b} \right)\]
\[ = 6 \left| \vec{a} \right|^2 + 2 \vec{a} . \vec{b} - 15 \vec{b} . \vec{a} - 5 \left| \vec{b} \right|^2 \]
\[ = 6 \left| \vec{a} \right|^2 + 2 \vec{a} . \vec{b} - 15 \vec{a} . \vec{b} - 5 \left| \vec{b} \right|^2 ( \vec{a} . \vec{b} = \vec{b} . \vec{a)} \]
\[ = 6 \left| \vec{a} \right|^2 - 13 \vec{a} . \vec{b} - 5 \left| \vec{b} \right|^2 \]
\[ = 6\left( 1 \right) - 13 \left( \frac{1}{2} \right) - 5\left( 1 \right) ...........\left[ \text{ From } (1) \right]\]
\[ = 1 - \frac{13}{2}\]
\[ = \frac{- 11}{2}\]
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