Math topic中文

Sum difference

use of sum and difference of two vectors to express given vectors in terms of two coplanar vectors

Practice questions

  1. 1In the diagram, OA→=6a\overrightarrow{OA} = 6a, OB→=6b\overrightarrow{OB} = 6b and 3AN→=AB→3\overrightarrow{AN} = \overrightarrow{AB}.…
  2. 2In the diagram, OABC is a parallelogram and D is the midpoint of BC. BE and OC produced intersect at point F. It is give…
  3. 3ABCD is a parallelogram. A is the point (−6,−2)(-6,-2) and B is the point (−4,−7)(-4,-7). BC→=(8−2)\overrightarrow{BC} = \begin{pmatrix}8\\-2\end{pmatrix}…
  4. 4It is given that PQ→=(−56)\overrightarrow{PQ}=\begin{pmatrix}-5\\6\end{pmatrix} and the position vector of PP is (2−4)\begin{pmatrix}2\\-4\end{pmatrix}…
  5. 5In the diagram, OA‾=12 AP‾\overline{OA}=\tfrac{1}{2}\,\overline{AP} and OB‾=14 OQ‾\overline{OB}=\tfrac{1}{4}\,\overline{OQ}. M is the mi…
  6. 6CBEO is a parallelogram. The position vectors of C and E, relative to O, are a\mathbf{a} and b\mathbf{b} respectively.…
  7. 7A is the point (2,3) and B is the point (3,-5). Find the equation of the line AB. Answer ... Find the length of AB. Answ…
  8. 8OABC is a trapezium. OA→=a\overrightarrow{OA}=a, OC→=2c\overrightarrow{OC}=2c and CB→=2a\overrightarrow{CB}=2a. OP:PB=1:3OP : PB = 1 : 3.…
  9. 9OABD is a parallelogram. O is the origin. A is the point (-7, 3) and AB→=(412)\overrightarrow{AB}=\begin{pmatrix}4\\12\end{pmatrix}…
  10. 10ABC is a triangle. D is the point on BC such that 3BD = 2DC. Find AD in terms of m and n. Answer ... R is on AD produced…
  11. 11OBC, ADC, AEB and OED are straight lines. OA→=a\overrightarrow{OA}=a and OB→=b\overrightarrow{OB}=b. OB=13OCOB=\dfrac{1}{3}OC, AE:EB=3:2AE:EB=3:2…
  12. 12The position vector of A is (34)\begin{pmatrix}3\\4\end{pmatrix} and the position vector of B is (710)\begin{pmatrix}7\\10\end{pmatrix}…
  13. 13ORS is a triangle. J is a point on RS such that RJ:JS=3:2RJ:JS = 3:2. OR→=r\overrightarrow{OR} = r and OS→=s\overrightarrow{OS} = s.…
  14. 14The position vector of point X is (8−4)\begin{pmatrix}8\\-4\end{pmatrix}. The position vector of point Y is (64)\begin{pmatrix}6\\4\end{pmatrix}…
  15. 15In the diagram ABCD is a parallelogram. The diagonal AC and BD intersect at E. AF→=2p\overrightarrow{AF} = 2p and AB→=6p+4q\overrightarrow{AB} = 6p + 4q…
  16. 16In the diagram, OA→=a\overrightarrow{OA} = a and OB→=b\overrightarrow{OB} = b. 2OB→=3BC→,  OD→=3OA→,  and 2AX→=XB→.2\overrightarrow{OB} = 3\overrightarrow{BC},\; \overrightarrow{OD} = 3\overrightarrow{OA},\; \text{and } 2\overrightarrow{AX} = \overrightarrow{XB}.…
  17. 17PQR is a triangle. PQ→=(2−3)\overrightarrow{PQ}=\begin{pmatrix}2\\-3\end{pmatrix}, PR→=(−39)\overrightarrow{PR}=\begin{pmatrix}-3\\9\end{pmatrix}…
  18. 18AB→=(−39), BC→=(25), CD→=(d4).\overrightarrow{AB}=\begin{pmatrix}-3\\9\end{pmatrix},\ \overrightarrow{BC}=\begin{pmatrix}2\\5\end{pmatrix},\ \overrightarrow{CD}=\begin{pmatrix}d\\4\end{pmatrix}.…
  19. 19The position vector of point Q is (−7−3)\begin{pmatrix}-7\\-3\end{pmatrix} and the position vector of point R is (51)\begin{pmatrix}5\\1\end{pmatrix}…
  20. 20The figure shows a triangle ABC. E is the mid-point of AC. AB‾=3a\overline{AB}=3a and AC‾=3c\overline{AC}=3c. K lies on EB such…
  21. 21The figure shows a rectangle ABCD with AD = 15 cm. E is on DC produced such that DE = 8 cm. The area of shaded part X is…
  22. 22The points A, B and C have coordinates (1, 0), (3, 7) and (3, 5) respectively. Find ∠BAC\angle BAC. Answer ___
  23. 23Point A is translated to point B by the vector (−54)\begin{pmatrix}-5\\4\end{pmatrix}. The position vector of point B is (38)\begin{pmatrix}3\\8\end{pmatrix}…
  24. 24The position vectors of points A and B are (−93)\begin{pmatrix}-9\\3\end{pmatrix} and (1−12)\begin{pmatrix}1\\-12\end{pmatrix}…