VECTOR! help plz :D?

Question

O is the origin. A n B are points such that OA = 4i + 5j n OB = 6i + j. Point C is situated on the line AB. Given that OC= (2t+2)i + tj. find



a) find value of t

b) ratio AC:CB

Answer 1

AB = OB - OA = (6i + j) - (4i + 5j) = 2i - 4j

So the line thru AB is:

L1 = OA + sAB = <4, 5> + s<2, -4>

Any non-zero multiple of the vector will do as well.
Divide by 2.

L1 = OA + sAB = <4, 5> + s<1, -2>
where s is a scalar ranging over the real numbers

The line L1 is intersected by the line thru the origin and C. The equation of that line is:

L2 = <2t + 2, t>

L1 = L2

<4, 5> + t<1, -2> = <2t + 2, t>
(4 + s)i + (5 - 2s)j = (2 + 2t)i + tj

Now we have two equations in two unknowns.

4 + s = 2 + 2t
5 - 2s = t

Add twice the first equation to the second.

13 = 4 + 5t
9 = 5t
t = 9/5

C = (2t + 2, t) = [2(9/5) + 2, 9/5] = (28/5, 9/5)

We also have points A and B.

A = (4, 5)
B = (6, 1)

The ratio of AC:CB is proportional to the change of either x or y over the interval. Let's use y.

AC:CB = (Ay - Cy) : (Cy - By)
AC:CB = (5 - 9/5) : (9/5 - 1) = (16/5) : (4/5) = 4 : 1

Answer 2

USE SLOPES
draw a figure

the slopes must be same.
5-1=4
6-4=2
4/2=2

2=(5-2t-2)/(t-4)

t=11/4

use distance between to points for AC/CB

Answer 3

(4,5) and (6,1) ... line has slope -2 and eqn
y-1 = -2 ( x-6)
y-1 = -2x +12
y = -2x +13

x= 2t + 2 and y = t..plug in and solve
t = -2(2t+2) + 13
t = -4t - 4 +13
5t = 9
t=9/5

C = 28/5 i + 9/5 j
AC = <28/5 - 4, 9/5 - 5> = < 8/5 , -16/5 >
CB = <28/5 - 6 , 9/5 - 1> = < -2/5 , 4/5 >
IACI = ( sqrt320 )/ 5
ICBI = (sqrt20) / 5

ratio sqrt 320 : sqrt 20 = 8sqrt5 : 2 sqrt5 = 8:2 or 4:1
ratio AC:AB = 4:1

:)