Cho \(\Delta ABC,\) gọi \({A_1},\,\,{B_1},\,\,{C_1}\) lần lượt là trung điểm của \(BC,\,\,CA,\,\,AB.\)
1.Chứng minh rằng \(\overrightarrow {A{A_1}} + \overrightarrow {B{B_1}} + \overrightarrow {C{C_1}} = \overrightarrow 0 .\)
2.Đặt \(\overrightarrow {B{B_1}} = \overrightarrow u ,\,\,\,\overrightarrow {C{C_1}} = \overrightarrow v .\) Tính \(\overrightarrow {BC} ,\,\,\overrightarrow {CA} ,\,\,\overrightarrow {AB} \) theo \(\overrightarrow u ,\,\,\overrightarrow v .\)
A.\(\begin{array}{*{20}{l}}{\overrightarrow {BC} = \frac{2}{3}\vec u - \frac{2}{3}\vec v}\\{\overrightarrow {AB} = - \frac{4}{3}\vec u - \frac{2}{3}\vec v}\\{\overrightarrow {CA} = \frac{2}{3}\vec u + \frac{4}{3}\vec v}\end{array}\)
B.\(\begin{array}{*{20}{l}}{\overrightarrow {BC} = \frac{2}{3}\vec u - \frac{2}{3}\vec v}\\{\overrightarrow {AB} = \frac{4}{3}\vec u + \frac{2}{3}\vec v}\\{\overrightarrow {CA} = \frac{2}{3}\vec u + \frac{4}{3}\vec v}\end{array}\)
C.\(\begin{array}{l}\overrightarrow {BC} = \frac{2}{3}\overrightarrow u + \frac{2}{3}\overrightarrow v \\\overrightarrow {AB} = \frac{4}{3}\overrightarrow u - \frac{2}{3}\overrightarrow v \\\overrightarrow {CA} = \frac{2}{3}\overrightarrow u - \frac{4}{3}\overrightarrow v \end{array}\)
D.\(\begin{array}{l}\overrightarrow {BC} = - \frac{2}{3}\overrightarrow u - \frac{2}{3}\overrightarrow v \\\overrightarrow {AB} = \frac{4}{3}\overrightarrow u + \frac{2}{3}\overrightarrow v \\\overrightarrow {CA} = 2\overrightarrow u + \frac{4}{3}\overrightarrow v \end{array}\)
1.Chứng minh rằng \(\overrightarrow {A{A_1}} + \overrightarrow {B{B_1}} + \overrightarrow {C{C_1}} = \overrightarrow 0 .\)
2.Đặt \(\overrightarrow {B{B_1}} = \overrightarrow u ,\,\,\,\overrightarrow {C{C_1}} = \overrightarrow v .\) Tính \(\overrightarrow {BC} ,\,\,\overrightarrow {CA} ,\,\,\overrightarrow {AB} \) theo \(\overrightarrow u ,\,\,\overrightarrow v .\)
A.\(\begin{array}{*{20}{l}}{\overrightarrow {BC} = \frac{2}{3}\vec u - \frac{2}{3}\vec v}\\{\overrightarrow {AB} = - \frac{4}{3}\vec u - \frac{2}{3}\vec v}\\{\overrightarrow {CA} = \frac{2}{3}\vec u + \frac{4}{3}\vec v}\end{array}\)
B.\(\begin{array}{*{20}{l}}{\overrightarrow {BC} = \frac{2}{3}\vec u - \frac{2}{3}\vec v}\\{\overrightarrow {AB} = \frac{4}{3}\vec u + \frac{2}{3}\vec v}\\{\overrightarrow {CA} = \frac{2}{3}\vec u + \frac{4}{3}\vec v}\end{array}\)
C.\(\begin{array}{l}\overrightarrow {BC} = \frac{2}{3}\overrightarrow u + \frac{2}{3}\overrightarrow v \\\overrightarrow {AB} = \frac{4}{3}\overrightarrow u - \frac{2}{3}\overrightarrow v \\\overrightarrow {CA} = \frac{2}{3}\overrightarrow u - \frac{4}{3}\overrightarrow v \end{array}\)
D.\(\begin{array}{l}\overrightarrow {BC} = - \frac{2}{3}\overrightarrow u - \frac{2}{3}\overrightarrow v \\\overrightarrow {AB} = \frac{4}{3}\overrightarrow u + \frac{2}{3}\overrightarrow v \\\overrightarrow {CA} = 2\overrightarrow u + \frac{4}{3}\overrightarrow v \end{array}\)
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