Solution: The cosine of the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ is given by $\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$. Compute the dot product: $\mathbf{a} \cdot \mathbf{b} = (1)(0) + (0)(1) + (1)(1) = 1$. The magnitudes are $\|\mathbf{a}\| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2}$ and $\|\mathbf{b}\| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$. Thus, $\cos\theta = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$. The final answer is $\boxed{\dfrac{1}{2}}$.

Sun, 03 May 2026 12:30:22 +0000

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