Question

The vectors \[\vec{a} \text{ and } \vec{b}\] satisfy the equations \[2 \vec{a} + \vec{b} = \vec{p} \text{ and } \vec{a} + 2 \vec{b} = \vec{q} , \text{ where } \vec{p} = \hat{i} + \hat{j} \text{ and } \vec{q} = \hat{i} - \hat{j} .\] the angle between \[\vec{a} \text{ and } \vec{b}\] then 

Options

•  \[\cos \theta = \frac{4}{5}\]

•  \[\sin \theta = \frac{1}{\sqrt{2}}\]

•  \[\cos \theta = - \frac{4}{5}\]

•  \[\cos \theta = - \frac{3}{5}\] 

Answer 1

 \[\cos \theta = - \frac{4}{5}\]  

\[\text{ Given that }\]

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

\[ \vec{a} + 2 \vec{b} = \vec{q} . . . \left( 2 \right)\]

\[\text{ Solving these two we get }\]

\[ \vec{a} = \frac{2 \vec{p} - \vec{q}}{3}, \vec{b} = \frac{2 \vec{q} - \vec{p}}{3}\]

\[\text{ And we have }\]

\[ \vec{p} = \hat{i} + \hat{j} \text{ and } \vec{q} = \hat{i} - \hat{j} \]

\[\text{ Substituting the values of } \vec{p} \text{ and } \vec{q,} \text{ we get }\]

\[ \vec{a} = \frac{2 \vec{p} - \vec{q}}{3} = \frac{2\left( \hat{i} + \hat{j} \right) - \left( \hat{i}- \hat{j} \right)}{3} = \frac{\hat{i} + 3 \hat{j}}{3}\]

\[ \Rightarrow \left| \vec{a} \right| = \frac{1}{3}\sqrt{1 + 9} = \frac{\sqrt{10}}{3}\]

\[ \vec{b} = \frac{2 \vec{q} - \vec{p}}{3} = \frac{2\left( \hat{i} - \hat{j} \right) - \left( \hat{i} + \hat{j} \right)}{3} = \frac{\hat{i} - 3 \hat{j}}{3}\]

\[ \Rightarrow \left| \vec{b} \right| = \frac{1}{3}\sqrt{1 + 9} = \frac{\sqrt{10}}{3}\]

\[ \vec{a} . \vec{b} = \frac{1}{9} \left( 1 - 9 \right) = \frac{- 8}{9}\]

\[\text{ We know that }\]

\[ \vec{a} . \vec{b} = \left| \vec{a} \right| \left| \vec{b} \right| \cos \theta\]

\[ \Rightarrow \frac{- 8}{9} = \frac{\sqrt{10}}{3} \times \frac{\sqrt{10}}{3} \cos \theta\]

\[ \Rightarrow \frac{- 8}{9} = \frac{10}{9}\cos \theta\]

\[ \Rightarrow \cos \theta = \frac{- 8}{9} \times \frac{9}{10} = \frac{- 4}{5}\]

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