# Thread: vector question

1. ## vector question

OAB is a triangle with OA = a and OB = b; C is the mid-point of OB, D is the mid-point of AB and E is the mid-point of OA; OD and AC intersect at F.
If AF = hAC and OF = kOD show that h = k = 2/3
well ive got that
AC = 1/2(b - a)
and
OD = (1/2)b - (1/2)a

AF = h/2(b - a)
and
OF = k((1/2)b - (1/2)a)

and this what i think i need to do get the intersection
h/2(b - a) = k((1/2)b - (1/2)a)

but i dont know where to go from there ?

2. Originally Posted by renlok
well ive got that
AC = 1/2(b - a)
and
OD = (1/2)b - (1/2)a

AF = h/2(b - a)
and
OF = k((1/2)b - (1/2)a)

and this what i think i need to do get the intersection
h/2(b - a) = k((1/2)b - (1/2)a)

but i dont know where to go from there ?
1. Draw a sketch!

2. $\overrightarrow{AC} = -\vec a + \frac12 \vec b$

$\overrightarrow{OD} = \frac12 (\vec a + \vec b)$

3. $\overrightarrow{OF} = \vec a + h \cdot \overrightarrow{AC}$

$\overrightarrow{OF} = k \cdot \overrightarrow{OD}$

Consequently you get:

$k \cdot \frac12 (\vec a + \vec b) = \vec a + h \cdot \left(-\vec a +\frac12 \vec b \right)$

Expand and rearrange:

$\frac12 k \vec a + \frac12 k \vec b = (1-h) \vec a + \frac12 h \vec b$

4. That means you know now:

$\left|\begin{array}{rcl}\frac12 k&=& 1-h \\ \frac12 k &=& \frac12 h \end{array} \right.$

5. Solve this system of equations for h, k.

3. oh so you have to get if for the same vector i get it now