Câu 1.
Theo định lí Pitago ta có: \(BC = \sqrt {A{B^2} + A{C^2}} = \sqrt {36 + 64} = 10\)
Ta có \(\overrightarrow {GB} - \overrightarrow {GC} = \overrightarrow {CB} \) . Suy ra \(\left| {\overrightarrow {GB} - \overrightarrow {GC} } \right| = \left| {\overrightarrow {CB} } \right| = CB = 10\)
Gọi M là trung điểm BC. Ta có \(\overrightarrow {GB} + \overrightarrow {GC} = 2\overrightarrow {GM} \) .
Mà \(GM = \dfrac{1 }{3}AM = \dfrac{1 }{6}BC = \dfrac{10} {6} = \dfrac{5 }{ 3}\)
Vậy \(\left| {\overrightarrow {GB} + \overrightarrow {GC} } \right| = \left| {2\overrightarrow {GM} } \right| = 2GM = \dfrac{10}{3}\)
Câu 2.
Ta có
\(\overrightarrow {BM} + \overrightarrow {DP} \)
\(= \overrightarrow {AM} - \overrightarrow {AB} + \overrightarrow {AP} - \overrightarrow {AD} \)
\( = \left( {\overrightarrow {AM} + \overrightarrow {AP} } \right) - \left( {\overrightarrow {AB} + \overrightarrow {AD} } \right) \)
\(= \overrightarrow {AN} - \overrightarrow {AC} = \overrightarrow {CN} \)
Câu 3.
Ta có:
\(\eqalign{ & \overrightarrow {BG} = \frac{1}{3}\left( {\overrightarrow {BO} + \overrightarrow {BC} + \overrightarrow {BD} } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{3}\left( {\overrightarrow {AO} - \overrightarrow {AB} + \overrightarrow {AC} - \overrightarrow {AB} + \overrightarrow {AD} - \overrightarrow {AB} } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{3}\left[ { - 3\overrightarrow {AB} + \overrightarrow {AC} + \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {AC} } \right) + \overrightarrow {AB} + \overrightarrow {AC} } \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{3}\left( { - \frac{3}{2}\overrightarrow {AB} + \frac{5}{2}\overrightarrow {AC} } \right) = - \frac{1}{2}\overrightarrow {AB} + \frac{5}{6}\overrightarrow {AC} {\text{ }} \cr} \)
