Page 277 - MATHEMATICS COURSE FOR SECONDARY SCHOOLS BOOK 2

Given the two end points, the position vector can also be found by calculation.
B(x2, y2) y2 - y,
The magnitude of PQ, PQ = √82 + (-3)2 = √64+9
= √73
= 8.54 units
The magnitude of MN, MN = √(-6)2 +(-4)2 = √36+16
= √52
= 7.21 units
A (x,y,)
Let A(x,y), be the start point and B(x2,y2) be the
end point of the displacement vector AB (ie AB).
then AB = ( 7 - 2 ) = ( 5 ) 4-1 3
x2 - x,
The magnitude of XY, XY = √72 + (-5)2 = √49+25
(x -x Then AB = y22-y11
= √74
= 8.60 units
EXERCISE 13.4
(a) A (3, 1) is the start point and B (9, 5) is the end point of AB. Find the displacement vector, AB, in the form.( xy )
(b) Calculate the magnitude of AB.
(a) P (4, 6) and Q (9, 8) are the start point and end point of PQ. Calculate the displacement vector, PQ, as a column vector.
(b) Evaluate the size of PQ.
(a) MN is formed by the points M (10, 5) and N (4, 8). Determine the displacement vector, MN, in the form ( xy )
(b) Determine the length of MN.
(a) The points X( 4, 9) and Y (12, 3) defines XY. Evaluate the displacement vector, XY.
)
(a) Given the end points, A (2, 1) and B (7, 4)
1.
2.
3.
4.
(b) Given the end points, P (2, 5) and Q (10, 2) then PQ = (10 - 2 )= (8 )
2-5 -3
(c) Given the end points, M (9, 6) and N (3, 2) then MN =(3 - 9 )= (-6 )
2 - 6 -4
(d) Given the end points, X (2, 7) and Y (9, 2) then XY = ( 9 - 2 ) = ( 7 )
The magnitude of vector AB, AB = √(x ,x )2+(y ,y )2 21 21
The magnitude of AB, AB = √52 + 32 = √ 25+9
= √34
= 5.83 units
2-7 -5
(b) Calculate the magnitude of XY. 271
AB =( x - x ) 21
y2 - y1