Trắc nghiệm Tổng và hiệu của hai vectơ có đáp án (Thông hiểu)

A. $\stackrel{\to }{AM}$

B. $\stackrel{\to }{PB}$

C. $\stackrel{\to }{AP}$

D. $\stackrel{\to }{MN}$

A. $\stackrel{\to }{AB}+\stackrel{\to }{AC}=\stackrel{\to }{BC}$

B. $\stackrel{\to }{MP}+\stackrel{\to }{NM}=\stackrel{\to }{NP}$

C. $\stackrel{\to }{CA}+\stackrel{\rightharpoonup }{BA}=\stackrel{\to }{CB}$

D. $\stackrel{\to }{AA}+\stackrel{\to }{BB}=\stackrel{\rightharpoonup }{AB}$

A. $\stackrel{\to }{AM}+\stackrel{\to }{MB}+\stackrel{\to }{BA}=\stackrel{\to }{0}$

B. $\stackrel{\to }{MA}+\stackrel{\to }{MB}=\stackrel{\to }{AB}$

C. $\stackrel{\to }{MA}+\stackrel{\to }{MB}=\stackrel{\to }{MC}$

D. $\stackrel{\to }{AB}+\stackrel{\to }{AC}=\stackrel{\to }{AM}$

A. $\left(\stackrel{\to }{AB}+\stackrel{\to }{AC}\right)=a\sqrt{2}$

B. $\left(\stackrel{\to }{AB}+\stackrel{\to }{AC}\right)=\frac{a\sqrt{2}}{2}$

C. $\left(\stackrel{\to }{AB}+\stackrel{\to }{AC}\right)=2a$

D. $\left(\stackrel{\to }{AB}+\stackrel{\to }{AC}\right)=a$

A. $\left(\stackrel{\to }{CA}+\stackrel{\to }{AB}\right)=2$

B. $\left(\stackrel{\to }{CA}+\stackrel{\to }{AB}\right)=2\sqrt{13}$

C. $\left(\stackrel{\to }{CA}+\stackrel{\to }{AB}\right)=5$

D. $\left(\stackrel{\to }{CA}+\stackrel{\to }{AB}\right)=\sqrt{13}$

A. $\stackrel{\to }{AM}$

B. $\stackrel{\to }{PB}$

C. $\stackrel{\to }{AP}$

D. $\stackrel{\to }{MN}$

A. $\left(\stackrel{\to }{OB}+\stackrel{\to }{OC}\right)=a$

B. $\left(\stackrel{\to }{OB}+\stackrel{\to }{OC}\right)=a\sqrt{2}$

C. $\left(\stackrel{\to }{OB}+\stackrel{\to }{OC}\right)=\frac{a}{2}$

D. $\left(\stackrel{\to }{OB}+\stackrel{\to }{OC}\right)=\frac{a\sqrt{2}}{2}$

A. $\stackrel{\to }{OA}+\stackrel{\to }{OC}+\stackrel{\to }{OE}=\stackrel{\to }{0}$

B. $\stackrel{\to }{BC}+\stackrel{\to }{FE}=\stackrel{\to }{AD}$

C. $\stackrel{\to }{OA}+\stackrel{\to }{OB}+\stackrel{\to }{OC}=\stackrel{\to }{EB}$

D. $\stackrel{\to }{AB}+\stackrel{\to }{CD}+\stackrel{\to }{FE}=\stackrel{\to }{0}$

A. a

B. $\sqrt{2}a$

C. $\frac{a}{2}$

D. 2a

A. $\stackrel{\to }{AB}+\stackrel{\to }{CD}+\stackrel{\to }{EF}$ $=\stackrel{\to }{AF}+\stackrel{\to }{ED}+\stackrel{\to }{BC}$

B. $\stackrel{\to }{AB}+\stackrel{\to }{CD}+\stackrel{\to }{EF}$ $=\stackrel{\to }{AF}+\stackrel{\to }{ED}+\stackrel{\to }{CB}$

C. $\stackrel{\to }{AE}+\stackrel{\to }{BF}+\stackrel{\to }{DC}=$ $\stackrel{\to }{DF}+\stackrel{\to }{BE}+\stackrel{\to }{AC}$

D. $\stackrel{\to }{AC}+\stackrel{\to }{BE}+\stackrel{\to }{EF}=$ $\stackrel{\to }{AD}+\stackrel{\to }{BF}+\stackrel{\to }{EC}$

A. $\left(\stackrel{\to }{AB}-\stackrel{\to }{DA}\right)=0$

B. $\left(\stackrel{\to }{AB}-\stackrel{\to }{DA}\right)=a$

C. $\left(\stackrel{\to }{AB}-\stackrel{\to }{DA}\right)=a\sqrt{2}$

D. $\left(\stackrel{\to }{AB}-\stackrel{\to }{DA}\right)=2a$

A. 2

B. 4

C. 8

D. $2\sqrt{3}$

A. $\stackrel{\to }{BC}+\stackrel{\to }{AB}$

B. $-\stackrel{\to }{OA}+\stackrel{\to }{OC}$

C. $\stackrel{\to }{BA}+\stackrel{\to }{DA}$

D. $\stackrel{\to }{DC}-\stackrel{\to }{CB}$

A. M là điểm sao cho tứ giác ABMC là hình bình hành

B. M là trọng tâm tam giác ABC

C. M là điểm sao cho tứ giác BAMC là hình bình hành

D. M thuộc trung trực của AB

A. $\stackrel{\to }{AC}=\stackrel{\to }{BD}$

B. $\stackrel{\to }{AB}+\stackrel{\to }{AC}+\stackrel{\to }{AD}=\stackrel{\to }{0}$

C. $\left(\stackrel{\to }{AB}-\stackrel{\to }{AD}\right)=\left(\stackrel{\to }{AB}+\stackrel{\to }{AD}\right)$

D. $\left(\stackrel{\to }{BC}+\stackrel{\to }{BD}\right)=\left(\stackrel{\to }{AC}-\stackrel{\to }{AB}\right)$