Trắc nghiệm Tích của vectơ với một số có đáp án (Thông hiểu)

A. $\stackrel{\to }{AI}=\frac{1}{4}\left(\stackrel{\rightharpoonup }{AB}+\stackrel{\rightharpoonup }{AC}\right)$

B. $\stackrel{\to }{AI}=\frac{1}{4}\left(\stackrel{\rightharpoonup }{AB}-\stackrel{\rightharpoonup }{AC}\right)$

C. $\stackrel{\to }{AI}=\frac{1}{4}\stackrel{\rightharpoonup }{AB}+\frac{1}{2}\stackrel{\rightharpoonup }{AC}$

D. $\stackrel{\to }{AI}=\frac{1}{4}\stackrel{\rightharpoonup }{AB}-\frac{1}{2}\stackrel{\rightharpoonup }{AC}$

A. $\stackrel{\to }{AG}=\frac{\stackrel{\to }{AB}+\stackrel{\to }{AC}}{2}$

B. $\stackrel{\to }{AG}=\frac{\stackrel{\to }{AB}+\stackrel{\to }{AC}}{3}$

C. $\stackrel{\to }{AG}=\frac{3(\stackrel{\to }{AB}+\stackrel{\to }{AC})}{2}$

D. $\stackrel{\to }{AG}=\frac{3(\stackrel{\to }{AB}+\stackrel{\to }{AC})}{3}$

A. $\stackrel{\to }{AG}=\frac{2}{3}\left(\stackrel{\to }{AB}+\stackrel{\to }{AC}\right)$

B. $\stackrel{\to }{AG}=\frac{1}{3}\left(\stackrel{\to }{AB}+\stackrel{\to }{AC}\right)$

C. $\stackrel{\to }{AG}=\frac{1}{3}\stackrel{\to }{AB}+\frac{2}{2}\stackrel{\to }{AC}$

D. $\stackrel{\to }{AG}=\frac{2}{3}\stackrel{\to }{AB}+3\stackrel{\to }{AC}$

A. $\stackrel{\to }{IB}+2\stackrel{\to }{IC}+\stackrel{\to }{IA}=\stackrel{\to }{0}$

B. $\stackrel{\to }{IB}+\stackrel{\to }{IC}+2\stackrel{\to }{IA}=\stackrel{\to }{0}$

C. $2\stackrel{\to }{IB}+\stackrel{\to }{IC}+\stackrel{\to }{IA}=\stackrel{\to }{0}$

D. $\stackrel{\to }{IB}+\stackrel{\to }{IC}+\stackrel{\to }{IA}=\stackrel{\to }{0}$

A. Nếu $\stackrel{\to }{AB}=\stackrel{\to }{AC}thì\left(\stackrel{\to }{AB}\right)=\left(\stackrel{\to }{AC}\right)$

B. $\stackrel{\to }{AB}=\stackrel{\to }{CD}$ thì A, B, C, D thẳng hàng

C. Nếu $3\stackrel{\to }{AB}+7\stackrel{\to }{AC}=\stackrel{\to }{0}$ thì A, B, C thẳng hàng

D. $\stackrel{\to }{AB}-\stackrel{\to }{CD}=\stackrel{\to }{DC}-\stackrel{\to }{BA}$

A. $2\stackrel{\to }{a}+\stackrel{\to }{b},\stackrel{\to }{a}+2\stackrel{\to }{b}$

B. $2\stackrel{\to }{a}-\stackrel{\to }{b},\stackrel{\to }{a}-2\stackrel{\to }{b}$

C. $5\stackrel{\to }{a}+\stackrel{\to }{b},-10\stackrel{\to }{a}-2\stackrel{\to }{b}$

D. $\stackrel{\to }{a}+\stackrel{\to }{b},\stackrel{\to }{a}-\stackrel{\to }{b}$

A. S=2a

B. S=a

C. $S=a\sqrt{3}$

D. $S=a\sqrt{2}$

A. a

B. $\left(1+\sqrt{2}\right)a$

C. $a\sqrt{5}$

D. $2a\sqrt{2}$

A. $\frac{1}{2}$

B. $-\frac{3}{2}$

C. $-\frac{1}{2}$

D. $\frac{3}{2}$

A. Ba điểm C, M, B thẳng hàng

B. AM là phân giác trong của góc $\widehat{BAC}$

C. A, M và trọng tâm tam giác ABC thẳng hàng

D. $\stackrel{\to }{AM}+\stackrel{\to }{BC}=\stackrel{\to }{0}$

A. $\left(\frac{3}{5};\frac{2}{5}\right)$

B. $\left(\frac{2}{5};\frac{3}{5}\right)$

C. $\left(-\frac{3}{5};\frac{2}{5}\right)$

D. $\left(\frac{3}{5};-\frac{2}{5}\right)$

A. $\stackrel{\to }{AM}=\frac{1}{4}\stackrel{\to }{AB}+3\stackrel{\to }{AC}$

B. $\stackrel{\to }{AM}=\frac{1}{4}\stackrel{\to }{AB}+\frac{3}{4}\stackrel{\to }{AC}$

C. $\stackrel{\to }{AM}=\frac{1}{4}\stackrel{\to }{AB}+\frac{1}{6}\stackrel{\to }{AC}$

D. $\stackrel{\to }{AM}=\frac{1}{2}\stackrel{\to }{AB}+\frac{1}{6}\stackrel{\to }{AC}$

A. $\stackrel{\to }{IM}=\frac{1}{6}\stackrel{\rightharpoonup }{AB}-\frac{1}{3}\stackrel{\to }{AC}$

B. $\stackrel{\to }{IM}=\frac{1}{6}\stackrel{\rightharpoonup }{AB}+\frac{1}{3}\stackrel{\to }{AC}$

C. $\stackrel{\to }{IM}=\frac{1}{3}\stackrel{\rightharpoonup }{AB}+\frac{1}{3}\stackrel{\to }{AC}$

D. $\stackrel{\to }{IM}=\frac{1}{3}\stackrel{\rightharpoonup }{AB}+\frac{1}{6}\stackrel{\to }{AC}$