If $O$ is the origin and if the coordinates of any two points $Q_{1}$ and $Q_{2}$ are $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ respectively, prove that $O Q_{1} . O Q_{2} cos ∠ Q_{1} O Q_{2} = x_{1} x_{2} + y_{1} y_{2}$

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Solution

By triangle law,
${(Q_{1} Q_{2})}^{2} = {(O Q_{1})}^{2} + {(O Q_{2})}^{2} - 2 O Q_{1} \cdot O Q_{2} cos (∠ Q_{1} O Q_{2})$

${(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2} = x_{1}^{2} + y_{1}^{2} + x_{2}^{2} + y_{2}^{2} - 2 O Q_{1} \cdot O Q_{2} cos (∠ Q_{1} O Q_{2})$

$- 2 (x_{1} x_{2} + y_{1} y_{2}) = - 2 O Q_{1} \cdot O Q_{2} cos (∠ Q_{1} O Q_{2})$

$O Q_{1} \cdot O Q_{2} cos ∠ Q_{1} O Q_{2} = x_{1} x_{2} + y_{1} y_{2}$

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