A) 0
B) \[\pi /2\]
C) 1
D) \[\pi /4\]
Correct Answer: C
Solution :
\[|\mathbf{x}-\mathbf{y}{{|}^{2}}=(\mathbf{x}-\mathbf{y})\,.\,(\mathbf{x}-\mathbf{y})=1+1-2|\mathbf{x}||\mathbf{y}|\cos \pi \] =\[2-2\,\cos \pi ,\,\,\,\therefore \text{ }|\mathbf{x}-\mathbf{y}|{{\,}^{2}}\,=4\] So, \[\frac{1}{2}|\mathbf{x}-\mathbf{y}|\,=1\], \[[\because \,\,|\mathbf{x}{{|}^{2}}=\,|\mathbf{y}{{|}^{2}}=1,|\mathbf{x}|\,=\,|\mathbf{y}|=1]\].
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