In the diagram below, QABC is a parallelogram, QA = a and QC = b. N is a point on QA such that QN : NA = 1:3. Find: a) DA b) BN in terms of a and b c) The lines AC and BN intersect at X. AX = k AC and BX = r BN. Express AX in terms of a and b d) Express AX in terms of a and b

Katika parallelogram OABC, vekta $\vec{OC}$ ni sawa na vekta $\vec{AB}$.

$$\vec{OC} = \vec{AB}$$ $$\vec{OC} = \mathbf{b}$$

Vekta $\vec{OC}$ ni $\boxed{\mathbf{b}}$.

Pointi N iko kwenye OA na ON : NA = 1 : 3. Hii inamaanisha $\vec{ON}$ ni sehemu ya $\vec{OA}$.

$$\vec{ON} = \frac{1}{1+3} \vec{OA} = \frac{1}{4} \vec{OA}$$ $$\vec{ON} = \frac{1}{4}\mathbf{a}$$

Tumia njia kutoka O kwenda B.

$$\vec{OB} = \vec{OA} + \vec{AB}$$ $$\vec{OB} = \mathbf{a} + \mathbf{b}$$

Tumia njia kutoka B kwenda N.

$$\vec{BN} = \vec{BO} + \vec{ON}$$

Kumbuka $\vec{BO} = -\vec{OB}$.

$$\vec{BN} = -(\mathbf{a} + \mathbf{b}) + \frac{1}{4}\mathbf{a}$$ $$\vec{BN} = -\mathbf{a} - \mathbf{b} + \frac{1}{4}\mathbf{a}$$ $$\vec{BN} = \left(1 - \frac{1}{4}\right)\mathbf{a} - \mathbf{b}$$ $$\vec{BN} = \boxed{-\frac{3}{4}\mathbf{a} - \mathbf{b}}$$

$$\vec{AC} = \vec{AO} + \vec{OC}$$

Kumbuka $\vec{AO} = -\vec{OA}$.

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

Tunajua $\vec{AX} = h\vec{AC}$.

$$\vec{OX} = \vec{OA} + \vec{AX}$$ $$\vec{OX} = \vec{OA} + h\vec{AC}$$ $$\vec{OX} = \mathbf{a} + h(-\mathbf{a} + \mathbf{b})$$ $$\vec{OX} = \mathbf{a} - h\mathbf{a} + h\mathbf{b}$$ $$\vec{OX} = (1-h)\mathbf{a} + h\mathbf{b} \quad \text{(Mlinganyo 1)}$$

Tunajua $\vec{BX} = k\vec{BN}$.

$$\vec{OX} = \vec{OB} + \vec{BX}$$ $$\vec{OX} = \vec{OB} + k\vec{BN}$$

Tulipata $\vec{OB} = \mathbf{a} + \mathbf{b}$ na $\vec{BN} = -\frac{3}{4}\mathbf{a} - \mathbf{b}$.

$$\vec{OX} = (\mathbf{a} + \mathbf{b}) + k\left(-\frac{3}{4}\mathbf{a} - \mathbf{b}\right)$$ $$\vec{OX} = \mathbf{a} + \mathbf{b} - \frac{3}{4}k\mathbf{a} - k\mathbf{b}$$ $$\vec{OX} = \left(1 - \frac{3}{4}k\right)\mathbf{a} + (1-k)\mathbf{b} \quad \text{(Mlinganyo 2)}$$

Kwa $\mathbf{a}$:

$$1-h = 1 - \frac{3}{4}k$$ $$h = \frac{3}{4}k \quad \text{(Mlinganyo 3)}$$

Kwa $\mathbf{b}$:

$$h = 1-k \quad \text{(Mlinganyo 4)}$$

Badilisha Mlinganyo 4 kwenye Mlinganyo 3:

$$1-k = \frac{3}{4}k$$ $$1 = k + \frac{3}{4}k$$ $$1 = \frac{4k+3k}{4}$$ $$1 = \frac{7}{4}k$$ $$k = \frac{4}{7}$$

Sasa pata $h$ kwa kutumia $h = 1-k$:

$$h = 1 - \frac{4}{7}$$ $$h = \frac{7-4}{7}$$ $$h = \frac{3}{7}$$

Thamani za $h$ na $k$ ni $\boxed{h = \frac{3}{7}, k = \frac{4}{7}}$.

$$\vec{OX} = (1-h)\mathbf{a} + h\mathbf{b}$$

Badilisha $h = \frac{3}{7}$:

$$\vec{OX} = \left(1-\frac{3}{7}\right)\mathbf{a} + \frac{3}{7}\mathbf{b}$$ $$\vec{OX} = \left(\frac{7-3}{7}\right)\mathbf{a} + \frac{3}{7}\mathbf{b}$$ $$\vec{OX} = \boxed{\frac{4}{7}\mathbf{a} + \frac{3}{7}\mathbf{b}}$$