# Thread: Vectors and point on the line segment.. (3d)

1. ## Vectors and point on the line segment.. (3d)

Two points A and B have position vectors (4i, -1j, 2k) and (1i, 5j, 3k). C is the point on the line segment AB such that AC / CB = 2. Find:

b) the displacement vector AC.

Please help me finding this displacement vector. I do know that it should be c - a or OC - OA but how would the rule fit in here.

2. For AC/CB to be 2, C has to be placed at 2/3 of the line segment AB. The parametric equation for the line passing through AB is:

$\displaystyle (x,y,z)=A+tAB$ where t is the parameter. This gives the equation:

$\displaystyle (x,y,z)=(4,-1,2)+t(-3,6,1)$

At t=0, we get A and at t=1, we get B. Therefore, by setting t=2/3, we'll get our point C:

C=(4,-1,2)+(2/3)(-3,6,1)

Then, AC is given by:

AC=C-A=(4,-1,2)+(2/3)(-3,6,1)-(4,-1,2)=(2/3)(-3,6,1)

AC=(-2,4,2/3)

3. For AC/CB to be 2, C has to be placed at 2/3 of the line segment AB. The parametric equation for the line passing through AB is:
I don't understand why is it 2/3. How do I get it? I get it by solving algebraically but I need to understand how did you get it without using algebra at all.

Thanks

4. Originally Posted by struck
I don't understand why is it 2/3. How do I get it? I get it by solving algebraically but I need to understand how did you get it without using algebra at all.

Thanks
The ratio of the line segments is 2:1. This means that it is split into thirds (as 2+1 = 3), with 2 of the thirds in the first segment (AC) and 1 of the thirds in the second segment (CB). Therefore, the placement of C must be $\displaystyle \frac{2}{3}$ of the line segment AB.