This is the only possible value under the constraint, so the maximum is:
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- \Rightarrow \|\mathbf{x}\|^2 + \|\mathbf{y}\|^2 + 2\mathbf{x} \cdot \mathbf{y} = 1
- Since $\|\mathbf{x}\| = \|\mathbf{y}\| = 1$, this becomes:
- + 1 + 2\mathbf{x} \cdot \mathbf{y} = 1 \Rightarrow 2 + 2\mathbf{x} \cdot \mathbf{y} = 1 \Rightarrow \mathbf{x} \cdot \mathbf{y} = -\frac{1}{2}
- \boxed{-\frac{1}{2}}
- Question:** A digital transformation algorithm rotates a vector in 3D space using $\tan(\theta)$, where $\theta$ is the angle between two planes. If $\cos \theta = \frac{1}{3}$, compute $\tan \theta$.
- Solution:** Use identity:
