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 \[\left( 2 \vec{a} - 5 \vec{b} \right) \cdot \left( 3 \vec{a} + \vec{b} \right) .\] 

Answer 1

\[\text{ We } have\]

\[\left| \vec{a} + \vec{b} \right| = \sqrt{3}\]

\[\text{ Squaring both sides , we get } \] 

\[ \left| \vec{a} + \vec{b} \right|^2 = 3\]

\[ \Rightarrow \left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 + 2 \vec{a} . \vec{b} = 3\]

\[ \Rightarrow 1 + 1 + 2 \vec{a} . \vec{b} = 3 ...........(\text{ Because } \vec{a} \text{ and } \vec{b} \text{ are  unit vectors) }\]

\[ \Rightarrow 2 + 2 \vec{a} . \vec{b} = 3\]

\[ \Rightarrow 2 \vec{a} . \vec{b} = 1\]

\[ \Rightarrow 2 \vec{a} . \vec{b} = 1\]

\[ \Rightarrow \vec{a} . \vec{b} = \frac{1}{2} . . . \left( 1 \right)\]

\[\text{ Now },\]

\[\left( 2 \vec{a} - 5 \vec{b} \right) . \left( 3 \vec{a} + \vec{b} \right)\]

\[ = 6 \left| \vec{a} \right|^2 + 2 \vec{a} . \vec{b} - 15 \vec{b} . \vec{a} - 5 \left| \vec{b} \right|^2 \]

\[ = 6 \left| \vec{a} \right|^2 + 2 \vec{a} . \vec{b} - 15 \vec{a} . \vec{b} - 5 \left| \vec{b} \right|^2 ( \vec{a} . \vec{b} = \vec{b} . \vec{a)} \]

\[ = 6 \left| \vec{a} \right|^2 - 13 \vec{a} . \vec{b} - 5 \left| \vec{b} \right|^2 \]

\[ = 6\left( 1 \right) - 13 \left( \frac{1}{2} \right) - 5\left( 1 \right) ...........\left[ \text{ From } (1) \right]\]

\[ = 1 - \frac{13}{2}\]

\[ = \frac{- 11}{2}\]