Cho ba vectơ \(\overrightarrow u (3;7;0),\overrightarrow v (2;3;1),\overrightarrow {\rm{w}} (3; - 2;4).\)
LG a
Chứng minh \(\overrightarrow u ,\overrightarrow v ,\overrightarrow {\rm{w}} \) không đồng phẳng.
Lời giải chi tiết:
\(\eqalign{ &\;\left[ {\overrightarrow u ,\overrightarrow v } \right] = \left( {\left| \matrix{ 7 \hfill \cr 3 \hfill \cr} \right.\left. \matrix{ 0 \hfill \cr 1 \hfill \cr} \right|;\left| \matrix{ 0 \hfill \cr 1 \hfill \cr} \right.\left. \matrix{ 3 \hfill \cr 2 \hfill \cr} \right|;\left| \matrix{ 3 \hfill \cr 2 \hfill \cr} \right.\left. \matrix{ 7 \hfill \cr 3 \hfill \cr} \right|} \right)\cr& = (7; - 3; - 5) \cr & \Rightarrow \left[ {\overrightarrow u ,\overrightarrow v } \right].\overrightarrow {\rm{w}} = 21 + 6 - 20 = 7 \ne 0. \cr} \)
Vậy \(\overrightarrow u ,\overrightarrow v ,\overrightarrow {\rm{w}} \) không đồng phẳng.
LG b
Biểu thị vec tơ \(\overrightarrow a ( - 4; - 12;3)\) theo ba vectơ \(\overrightarrow u ,\overrightarrow v ,\overrightarrow {\rm{w}} \).
Lời giải chi tiết:
\(\eqalign{\;\overrightarrow a = m\overrightarrow u + n\overrightarrow v + k\overrightarrow {\rm{w}} \cr & \Leftrightarrow \left\{ \matrix{ 3m + 2n + 3k = - 4 \hfill \cr 7m + 3n - 2k = - 12 \hfill \cr n + 4k = 3 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ m = - 5 \hfill \cr n = 7 \hfill \cr k = - 1. \hfill \cr} \right. \cr} \)
Vậy \(\overrightarrow a = - 5\overrightarrow u + 7\overrightarrow v - \overrightarrow {\rm{w}} .\)
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