Orthogonal Vectors Cosine
For certain vectors $ \mathbf{p} $ and $ \mathbf{q}, $ the vectors $ 3 \mathbf{p} + \mathbf{q} $ and $ 5 \mathbf{p} - 3 \mathbf{q} $ are orthogonal. Also, the vectors $ 2 \mathbf{p} + \mathbf{q} $ and $ 4 \mathbf{p} - 2 \mathbf{q} $ are orthogonal. If $ \theta $ is the angle between $ \mathbf{p} $ and $ \mathbf{q}, $ then find $ \cos \theta $.
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Solution
Since $ 2 \mathbf{p} + \mathbf{q} $ and $ 4 \mathbf{p} - 2 \mathbf{q} $ are orthogonal, $ (2 \mathbf{p} + \mathbf{q}) \cdot (4 \mathbf{p} - 2 \mathbf{q}) = 0 $. Expanding, we get
\[8 \mathbf{p} \cdot \mathbf{p} - 2 \mathbf{q} \cdot \mathbf{q} = 0,\]so $ \|\mathbf{q}\|^2 = 4 \|\mathbf{p}\|^2, $ and $ \|\mathbf{q}\| = 2 \|\mathbf{p}\| $.
Since $ 3 \mathbf{p} + \mathbf{q} $ and $ 5 \mathbf{p} - 3 \mathbf{q} $ are orthogonal, $ (3 \mathbf{p} + \mathbf{q}) \cdot (5 \mathbf{p} - 3 \mathbf{q}) = 0 $. Expanding, we get
\[15 \mathbf{p} \cdot \mathbf{p} - 4 \mathbf{p} \cdot \mathbf{q} - 3 \mathbf{q} \cdot \mathbf{q} = 0.\]Since $ \mathbf{q} \cdot \mathbf{q} = 4 \mathbf{p} \cdot \mathbf{p}, $
\[4 \mathbf{p} \cdot \mathbf{q} = 3 \mathbf{p} \cdot \mathbf{p}.\]Then
\[\cos \theta = \frac{\mathbf{p} \cdot \mathbf{q}}{\|\mathbf{p}\| \|\mathbf{q}\|} = \frac{\frac{3}{4} \mathbf{p} \cdot \mathbf{p}}{2 \|\mathbf{p}\|^2} = \boxed{\frac{3}{8}}.\]
Freeflows