# Trong không gian với hệ tọa độ Oxyz, cho $\stackrel{\to }{a}=\left(2\text{\hspace{0.17em}};\text{\hspace{0.17em}}-3\text{\hspace{0.17em}};\text{\hspace{0.17em}}3\right)$ , $\stackrel{\to }{b}=\left(0\text{\hspace{0.17em}};\text{\hspace{0.17em}}2\text{\hspace{0.17em}};-1\right)$ , $\stackrel{\to }{c}=\left(3\text{\hspace{0.17em}};-1\text{\hspace{0.17em}};5\right)$ . Tìm tọa độ của vectơ $\stackrel{\to }{u}=2\stackrel{\to }{a}+3\stackrel{\to }{b}-2\stackrel{\to }{c}$ .

A.$\left(10\text{\hspace{0.17em}};\text{\hspace{0.17em}}-2\text{\hspace{0.17em}};\text{\hspace{0.17em}}13\right)$ .
B.$\left(-2\text{\hspace{0.17em}};\text{\hspace{0.17em}}2\text{\hspace{0.17em}};\text{\hspace{0.17em}}-7\right)$ .
C.$\left(-2\text{\hspace{0.17em}};\text{\hspace{0.17em}}-2\text{\hspace{0.17em}};7\right)$ .
D.$\left(-2\text{\hspace{0.17em}};\text{\hspace{0.17em}}2\text{\hspace{0.17em}};\text{\hspace{0.17em}}7\right)$ .
Đáp án:B
Lời giải:Lời giải
Chn B
Ta có $\begin{array}{l}2\stackrel{\to }{a}=\left(4\text{\hspace{0.17em}};\text{\hspace{0.17em}}-6\text{\hspace{0.17em}};\text{\hspace{0.17em}}6\right)\\ 3\stackrel{\to }{b}=\left(0\text{\hspace{0.17em}};\text{\hspace{0.17em}}6\text{\hspace{0.17em}};-3\right)\\ 2\stackrel{\to }{c}=\left(6\text{\hspace{0.17em}};-2\text{\hspace{0.17em}};10\right)\end{array}}⇒2\stackrel{\to }{a}+3\stackrel{\to }{b}-2\stackrel{\to }{c}=\left(-2\text{\hspace{0.17em}};\text{\hspace{0.17em}}2\text{\hspace{0.17em}};-7\right)$ .

Vậy đáp án đúng là B.