In the figure below OB=b;OC=3OB and OA = a. Given that OD = 1/3 OA and AN = 1/2 AC, CD and AB meet at M. Determine in terms of a and b (i) AB (ii) CD. Given that CM = K CD and AM = h AB. Determine the values of the scalars K and h. Show that O,M and N are collinear.

$$\vec{AB} = \vec{AO} + \vec{OB}$$

Since $\vec{AO} = -\vec{OA}$, we have:

$$\vec{AB} = -\vec{OA} + \vec{OB}$$

Substitute the given vectors:

$$\vec{AB} = -\mathbf{a} + \mathbf{b}$$ $$\vec{AB} = \mathbf{b} - \mathbf{a}$$

The final answer is $\boxed{\mathbf{b} - \mathbf{a}}$.

$$\vec{CD} = \vec{CO} + \vec{OD}$$

Since $\vec{CO} = -\vec{OC}$, we have:

$$\vec{CD} = -\vec{OC} + \vec{OD}$$

Substitute the given vectors $\vec{OC} = 3\mathbf{b}$ and $\vec{OD} = \frac{1}{3}\vec{OA} = \frac{1}{3}\mathbf{a}$:

$$\vec{CD} = -3\mathbf{b} + \frac{1}{3}\mathbf{a}$$ $$\vec{CD} = \frac{1}{3}\mathbf{a} - 3\mathbf{b}$$

The final answer is $\boxed{\frac{1}{3}\mathbf{a} - 3\mathbf{b}}$.

From $\vec{CM} = K\vec{CD}$:

$$\vec{OM} - \vec{OC} = K(\vec{OD} - \vec{OC})$$ $$\vec{OM} = \vec{OC} + K\vec{OD} - K\vec{OC}$$ $$\vec{OM} = (1-K)\vec{OC} + K\vec{OD}$$

Substitute $\vec{OC} = 3\mathbf{b}$ and $\vec{OD} = \frac{1}{3}\mathbf{a}$:

$$\vec{OM} = (1-K)(3\mathbf{b}) + K\left(\frac{1}{3}\mathbf{a}\right)$$ $$\vec{OM} = \frac{K}{3}\mathbf{a} + (3-3K)\mathbf{b} \quad \text{(Equation 1)}$$

From $\vec{AM} = h\vec{AB}$:

$$\vec{OM} - \vec{OA} = h(\vec{OB} - \vec{OA})$$ $$\vec{OM} = \vec{OA} + h\vec{OB} - h\vec{OA}$$ $$\vec{OM} = (1-h)\vec{OA} + h\vec{OB}$$

Substitute $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$:

$$\vec{OM} = (1-h)\mathbf{a} + h\mathbf{b} \quad \text{(Equation 2)}$$

$$\frac{K}{3}\mathbf{a} + (3-3K)\mathbf{b} = (1-h)\mathbf{a} + h\mathbf{b}$$

Equating coefficients of $\mathbf{a}$:

$$\frac{K}{3} = 1-h \quad \text{(Equation 3)}$$

Equating coefficients of $\mathbf{b}$:

$$3-3K = h \quad \text{(Equation 4)}$$

Substitute Equation 4 into Equation 3:

$$\frac{K}{3} = 1 - (3-3K)$$ $$\frac{K}{3} = 1 - 3 + 3K$$ $$\frac{K}{3} = -2 + 3K$$

Multiply by 3:

$$K = -6 + 9K$$ $$6 = 8K$$ $$K = \frac{6}{8} = \frac{3}{4}$$

Substitute $K = \frac{3}{4}$ into Equation 4:

$$h = 3 - 3\left(\frac{3}{4}\right)$$ $$h = 3 - \frac{9}{4}$$ $$h = \frac{12}{4} - \frac{9}{4}$$ $$h = \frac{3}{4}$$

The final answer is $\boxed{K = \frac{3}{4}, h = \frac{3}{4}}$.

Using Equation 2 with $h = \frac{3}{4}$:

$$\vec{OM} = (1-\frac{3}{4})\mathbf{a} + \frac{3}{4}\mathbf{b}$$ $$\vec{OM} = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}$$

We are given $\vec{AN} = \frac{1}{2}\vec{AC}$.

$$\vec{ON} - \vec{OA} = \frac{1}{2}(\vec{OC} - \vec{OA})$$ $$\vec{ON} = \vec{OA} + \frac{1}{2}\vec{OC} - \frac{1}{2}\vec{OA}$$ $$\vec{ON} = (1-\frac{1}{2})\vec{OA} + \frac{1}{2}\vec{OC}$$

Substitute $\vec{OA} = \mathbf{a}$ and $\vec{OC} = 3\mathbf{b}$:

$$\vec{ON} = \frac{1}{2}\mathbf{a} + \frac{1}{2}(3\mathbf{b})$$ $$\vec{ON} = \frac{1}{2}\mathbf{a} + \frac{3}{2}\mathbf{b}$$

$$\vec{OM} = \frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}$$ $$\vec{ON} = \frac{1}{2}\mathbf{a} + \frac{3}{2}\mathbf{b}$$

We can factor out a scalar from $\vec{ON}$:

$$\vec{ON} = 2\left(\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\right)$$ $$\vec{ON} = 2\vec{OM}$$

Since $\vec{ON}$ is a scalar multiple of $\vec{OM}$ and both vectors originate from O, the points O, M, and N are collinear.

3 zimekamilika, zimebaki 2 leo. Unapiga hatua.