4. Relative to an origin 0, the position vectors of the points A and B are a and b respectively.
The position vectors of the points P, Q and R relative to 0 are 13a -2b, 22a+ b, and 10a -3b
respectively. (a) Show that the points P, Q and R are collinear. (b) Draw a diagram to show the
positions of P, Q and R relative to one other and evaluate
Solution
Given OP = 13 a - 2b;
OQ = 22a + b;
OR = 10a - 3b;
det ( 13 -2 1 ,
22 1 1,
10 -3 1) = 13(1 - (-3) ) - (-2)(22 - 10) + 1(22(-3) - 10) = 13(4) - (-24) + 1(-76) = 52 + 24 - 76 = 0
As det of the vectors = 0 , hence points are collinear.
PQ = OQ - OP = 22a + b - (13a - 2b) = 9a + 3b;
PR = OR - OP = 10a - 3b - (13a - 2b) = -3a - b ;
QR = OR - OQ = 10 a - 3b - (22a + b) = -12a - 4b; = 4( -3a -b) = 4 PR
Hence |PR |/|QR| = 1/4.
4. Relative to an origin 0, the position vectors of the points A and .pdf
1. 4. Relative to an origin 0, the position vectors of the points A and B are a and b respectively.
The position vectors of the points P, Q and R relative to 0 are 13a -2b, 22a+ b, and 10a -3b
respectively. (a) Show that the points P, Q and R are collinear. (b) Draw a diagram to show the
positions of P, Q and R relative to one other and evaluate
Solution
Given OP = 13 a - 2b;
OQ = 22a + b;
OR = 10a - 3b;
det ( 13 -2 1 ,
22 1 1,
10 -3 1) = 13(1 - (-3) ) - (-2)(22 - 10) + 1(22(-3) - 10) = 13(4) - (-24) + 1(-76) = 52 + 24 - 76 = 0
As det of the vectors = 0 , hence points are collinear.
PQ = OQ - OP = 22a + b - (13a - 2b) = 9a + 3b;
PR = OR - OP = 10a - 3b - (13a - 2b) = -3a - b ;
QR = OR - OQ = 10 a - 3b - (22a + b) = -12a - 4b; = 4( -3a -b) = 4 PR
Hence |PR |/|QR| = 1/4
