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Project \(\mathbfv\) onto \(\mathbfd\): - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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Multiply numerator and denominator by \(\sqrt{5} + 1\):
r = -rac{(5 - \sqrt{5})(\sqrt{5} + 1)}{(\sqrt{5} - 1)(\sqrt{5} + 1)}...
Thus, \( r = \sqrt{5} \) or \( r = -\sqrt{5} \). These correspond to p...
e 0\), and \(x
e y\) (to avoid division by zero in the original expression).
Check that \(x = \sqrt{5} y\):
Then \( rac{x+y}{x-y} = rac{\sqrt{5}+1}{\sqrt{5}-1} = rac{3+\sqrt{5...
Similarly for \(x = -\sqrt{5} y\):
\( rac{-\sqrt{5}+1}{-\sqrt{5}-1} = rac{\sqrt{5}-1}{\sqrt{5}+1} = ra...
Therefore, the solutions are all real pairs where \( x = \sqrt{5} y \)...

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