Question 3 (3 marks):
Diagram 3 shows vectors →AB, →AC and →AD−−→AB, −−→AC and −−→AD drawn on a square grid with sides of 1 unit.
Diagram 3
(a) Find |−→BA|.(b) Given →AB=b˜ and →AC=c˜,express in terms of b˜ and c˜ (i) →BC,(ii) →AD
Solution:
(a)
|−→BA|=√32+42=5 units
(b)(i)
→BC=→BA+→AC =−b˜+c˜ =c˜−b˜
(b)(ii)
→AD=→AB+→BD =b˜+2→BC =b˜+2(c˜−b˜) =2c˜−b˜
Diagram 3 shows vectors →AB, →AC and →AD−−→AB, −−→AC and −−→AD drawn on a square grid with sides of 1 unit.
Diagram 3(a) Find |−→BA|.(b) Given →AB=b˜ and →AC=c˜,express in terms of b˜ and c˜ (i) →BC,(ii) →AD
Solution:
(a)
|−→BA|=√32+42=5 units
(b)(i)
→BC=→BA+→AC =−b˜+c˜ =c˜−b˜
(b)(ii)
→AD=→AB+→BD =b˜+2→BC =b˜+2(c˜−b˜) =2c˜−b˜