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[br]\displaystyle[br]\text{Sphere } S \text{ intersects plane }\Pi. \\[br]S: x^2 + y^2 + (z-a)^2 = b^2\\[br]\Pi : z = 0 \\[br]\text{Solve simultaneously:} \\[br]x^2 + y^2 = b^2 - a^2. \\[br]\text{If } b^2 > a^2\text{, then } x^2 + y^2 = c^2 \longleftarrow \text{circle} \\[br]\text{If } b^2 = a^2\text{, then } x^2 + y^2 = 0 \longleftarrow \text{point } (0, 0, 0) \\[br]\text{If } b^2 < a^2\text{, then } x^2 + y^2 = -c^2 \longleftarrow \text{empty set.} \\ \\[br][br]\text{The vectors i, j and k are such that } \overrightarrow{AC} = 2\mathbf{i},\; \overrightarrow{AD} = 2\mathbf{k},\; \\[br]\text{and i, j and k form a right handed system of orthonormal base vectors.} \\[br]\overrightarrow{AC} = 2\mathbf{i}, \overrightarrow{AD} = 2\mathbf{k}, \overrightarrow{BE} = 3\mathbf{k}, \overrightarrow{CF} = 4\mathbf{k} \\[br]\overrightarrow{AB} = 2(\cos \frac{\pi}{3} \mathbf{i} + \sin \frac{\pi}{3} \mathbf{j}) = \mathbf{i} + \sqrt{3} \mathbf{j} \\[br]\overrightarrow{DE} = \overrightarrow{DA} + \overrightarrow{AB} + \overrightarrow{BE} = \begin{pmatrix} 1 \\ \sqrt{3} \\ 1 \end{pmatrix} \\[br]\text{Similarly, } \overrightarrow{EF} = \begin{pmatrix} 1 \\ -\sqrt{3} \\ 1 \end{pmatrix} \\[br]\overrightarrow{DE}\cdot \overrightarrow{EF} = \begin{pmatrix} 1 \\ \sqrt{3} \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ -\sqrt{3} \\ 1 \end{pmatrix} = -1 \\
[br]\displaystyle[br]\text{The pole } AD \text{ is shifted.} \\[br]\overrightarrow{AD} = \overrightarrow{AB} \cos \frac{\pi}{4} + 2\mathbf{k} \cos \frac{\pi}{4} \\[br]= \begin{pmatrix} 1/\sqrt{2} \\ \sqrt{3/2} \\ 1/\sqrt{2} \end{pmatrix}.[br]\overrightarrow{DE} = -\begin{pmatrix} 1/\sqrt{2} \\ \sqrt{3/2} \\ 1/\sqrt{2} \end{pmatrix} + \begin{pmatrix} 1 \\ \sqrt{3} \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ 3 \end{pmatrix} \\[br] = \frac{1}{\sqrt{2}} \begin{pmatrix} \sqrt{2} - 1 \\ \sqrt{3}(\sqrt{2} - 1) \\ 3\sqrt{2} - 1 \end{pmatrix}. \\ \\[br][br]\overrightarrow{DE} \cdot \overrightarrow{EF} = \frac{1}{\sqrt{2}} (\sqrt{2} - 1 - 3(\sqrt{2} - 1) + 3\sqrt{2} - 1) = 1 + \frac{1}{\sqrt{2}}.
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