Lee Ortiz
08/19/2023 · Elementary School
прямой DM , если \( \mathrm{D}(-3 ; 20 ;-4), \mathrm{M}(9 ; 24 ; 16) \). 6. Дан куб \( \mathrm{ABCDA} \mathrm{B}_{1} \mathrm{~B}_{1} \mathrm{D}_{1} \), ребро которого равно \( 1 \mathrm{cм} \). На диагонали AC его грани отметили точку F так, что \( \mathrm{AF}: \mathrm{FC}=3: 7 \). 1) Выразите вектор \( \overrightarrow{B_{1} F} \) через векторы \( \overrightarrow{B_{1} A_{1}}, \overrightarrow{B_{1} C_{1}} \) и \( \overrightarrow{B_{1} B} \). 2) Найдите модуль вектора \( \overrightarrow{B_{1} F} \).
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Quick Answer
1) Вектор \( \overrightarrow{B_1 F} \) можно выразить через векторы \( \overrightarrow{B_1 A_1}, \overrightarrow{B_1 C_1} \) и \( \overrightarrow{B_1 B} \) как:
\[
\overrightarrow{B_1 F} = -\frac{7}{10} \overrightarrow{B_1 A_1} + \frac{3}{10} \overrightarrow{B_1 C_1} - 1 \overrightarrow{B_1 B}
\]
2) Модуль вектора \( \overrightarrow{B_1 F} \) равен:
\[
|\overrightarrow{B_1 F}| = \frac{\sqrt{158}}{10} \, \text{см}
\]
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