IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: T.Q on April 22, 2010, 10:37:08 am
-
The point A, with coordinates (0, a, b) lies on the line l1, which has equation
r = 6i + 19j – k + t(i + 4j – 2k).
(a) Find the values of a and b.
(3)
The point P lies on l1 and is such that OP is perpendicular to l1, where O is the origin.
(b) Find the position vector of point P.
(6)
Given that B has coordinates (5, 15, 1),
(c) show that the points A, P and B are collinear and find the ratio AP : PB.
(4)
need help in part c with explanation plz
-
2 hours
-
The point A, with coordinates (0, a, b) lies on the line l1, which has equation
r = 6i + 19j – k + t(i + 4j – 2k).
(a) Find the values of a and b.
(3)
The point P lies on l1 and is such that OP is perpendicular to l1, where O is the origin.
(b) Find the position vector of point P.
(6)
Given that B has coordinates (5, 15, 1),
(c) show that the points A, P and B are collinear and find the ratio AP : PB.
(4)
Form the x cooridate6+t=0 so t=-6
a=19+4t=19+4*-6=-5
b=-1-2t=-1-2*-6=11
b)OP perpendicular to l so OP.v=0 where v is tangent vector of l v=i + 4j – 2k
P is on l so P=(6+t)i+(19+4t)j+(-1-2t)j
OP.v=6+t+4(19+4t)-2(-1-2t)=0 so 84+21t=-4
P=(6-4)i+(19+4*-4)j+(-1-2*-4)k=2i+3j+7k
-
thanx
but i need help in part c
-
AP=P-A =(2,3,7)-(0,-5,11)=(2,8,-4)
PB=B-P=(5,15,1)-(2,3,7)=(3,12,,-6)
1.5*AP=PB so AP:PB is 1:1.5 or 2:3
-
thanx :)
+rep
-
No probs. Will do a page on my website too