C4 vectors help

[T.Q]

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

[astarmathsandphysics]

2 hours

[astarmathsandphysics]

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

[T.Q]

thanx

but i need help in part c

[astarmathsandphysics]

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

[T.Q]

thanx


+rep

[astarmathsandphysics]

No probs. Will do a page on my website too