In the trapezium shown below PQ = 3ST. T divides SR in the ratio 4:1 and U is the midpoint of QT. PU and QR intersect at X. PX = hPU and QX = kQR.

Huu ni swali la vekta. Tutatumia taarifa zilizotolewa na sheria za vekta kutatua kila sehemu.

Taarifa Zilizotolewa:

• Trapezium PQRS.
• $\vec{PQ} = 3\vec{ST}$. Hii inamaanisha $\vec{PQ}$ ni sambamba na $\vec{ST}$ na urefu wake ni mara 3 ya $\vec{ST}$. Kwa kuwa T iko kwenye SR, hii inamaanisha $\vec{PQ}$ ni sambamba na $\vec{SR}$.
• T inagawanya SR kwa uwiano wa 4:1. Hivyo $\vec{ST} = \frac{4}{5}\vec{SR}$ na $\vec{TR} = \frac{1}{5}\vec{SR}$.
• U ni katikati ya QT. Hivyo $\vec{QU} = \vec{UT} = \frac{1}{2}\vec{QT}$.
• PU na QR zinakutana kwenye X. Hivyo X iko kwenye PU na X iko kwenye QR.
• $\vec{PX} = h\vec{PU}$ na $\vec{QX} = k\vec{QR}$.
• $\vec{PQ} = \mathbf{q}$ na $\vec{PS} = \mathbf{p}$.

Hatua za Kwanza:
Kutoka $\vec{PQ} = 3\vec{ST}$:

$$\vec{ST} = \frac{1}{3}\vec{PQ} = \frac{1}{3}\mathbf{q}$$

Kutoka T inagawanya SR kwa uwiano 4:1, na $\vec{ST} = \frac{4}{5}\vec{SR}$:

$$\vec{SR} = \frac{5}{4}\vec{ST} = \frac{5}{4}\left(\frac{1}{3}\mathbf{q}\right) = \frac{5}{12}\mathbf{q}$$

a) Eleza $\vec{QR}$ kwa kutumia $\mathbf{p}$ na $\mathbf{q}$.
Tunatumia njia ya vekta kutoka Q hadi R:

$$\vec{QR} = \vec{QP} + \vec{PS} + \vec{SR}$$

Tunajua $\vec{QP} = -\vec{PQ} = -\mathbf{q}$.

$$\vec{QR} = -\mathbf{q} + \mathbf{p} + \frac{5}{12}\mathbf{q}$$ $$\vec{QR} = \mathbf{p} - \frac{7}{12}\mathbf{q}$$ $$\boxed{\vec{QR} = \mathbf{p} - \frac{7}{12}\mathbf{q}}$$

b) Eleza $\vec{PX}$ kwa kutumia $\mathbf{p}$, $\mathbf{q}$ na $h$.
X iko kwenye PU, na $\vec{PX} = h\vec{PU}$. Kwanza tunahitaji kupata $\vec{PU}$.

$$\vec{PU} = \vec{PS} + \vec{SU}$$

Ili kupata $\vec{SU}$, tunahitaji $\vec{ST}$ na $\vec{TU}$.

$$\vec{SU} = \vec{ST} + \vec{TU}$$

U ni katikati ya QT, hivyo $\vec{TU} = \frac{1}{2}\vec{TQ} = -\frac{1}{2}\vec{QT}$.
Kwanza, pata $\vec{QT}$:

$$\vec{QT} = \vec{QP} + \vec{PS} + \vec{ST}$$ $$\vec{QT} = -\mathbf{q} + \mathbf{p} + \frac{1}{3}\mathbf{q}$$ $$\vec{QT} = \mathbf{p} - \frac{2}{3}\mathbf{q}$$

Sasa, pata $\vec{TU}$:

$$\vec{TU} = -\frac{1}{2}\left(\mathbf{p} - \frac{2}{3}\mathbf{q}\right) = -\frac{1}{2}\mathbf{p} + \frac{1}{3}\mathbf{q}$$

Sasa, pata $\vec{SU}$:

$$\vec{SU} = \frac{1}{3}\mathbf{q} + \left(-\frac{1}{2}\mathbf{p} + \frac{1}{3}\mathbf{q}\right) = -\frac{1}{2}\mathbf{p} + \frac{2}{3}\mathbf{q}$$

Sasa, pata $\vec{PU}$:

$$\vec{PU} = \mathbf{p} + \left(-\frac{1}{2}\mathbf{p} + \frac{2}{3}\mathbf{q}\right) = \frac{1}{2}\mathbf{p} + \frac{2}{3}\mathbf{q}$$

Mwishowe, pata $\vec{PX}$:

$$\vec{PX} = h\vec{PU} = h\left(\frac{1}{2}\mathbf{p} + \frac{2}{3}\mathbf{q}\right)$$ $$\boxed{\vec{PX} = \frac{h}{2}\mathbf{p} + \frac{2h}{3}\mathbf{q}}$$

c) Eleza $\vec{PX}$ kwa kutumia $\mathbf{p}$, $\mathbf{q}$ na $k$.
X iko kwenye QR, na $\vec{QX} = k\vec{QR}$.
Tunatumia njia ya vekta kutoka P hadi X:

$$\vec{PX} = \vec{PQ} + \vec{QX}$$ $$\vec{PX} = \vec{PQ} + k\vec{QR}$$

Tunajua $\vec{PQ} = \mathbf{q}$ na kutoka sehemu (a), $\vec{QR} = \mathbf{p} - \frac{7}{12}\mathbf{q}$.

$$\vec{PX} = \mathbf{q} + k\left(\mathbf{p} - \frac{7}{12}\mathbf{q}\right)$$ $$\vec{PX} = \mathbf{q} + k\mathbf{p} - \frac{7k}{12}\mathbf{q}$$ $$\vec{PX} = k\mathbf{p} + \left(1 - \frac{7k}{12}\right)\mathbf{q}$$ $$\boxed{\vec{PX} = k\mathbf{p} + \left(1 - \frac{7k}{12}\right)\mathbf{q}}$$

d) Pata thamani za $h$ na $k$.
Tunalinganisha milinganyo miwili ya $\vec{PX}$ kutoka sehemu (b) na (c):

$$\frac{h}{2}\mathbf{p} + \frac{2h}{3}\mathbf{q} = k\mathbf{p} + \left(1 - \frac{7k}{12}\right)\mathbf{q}$$

Kwa kuwa $\mathbf{p}$ na $\mathbf{q}$ si vekta sambamba, coefficients zao lazima ziwe sawa.
Linganisha coefficients za $\mathbf{p}$:

$$\frac{h}{2} = k \quad (1)$$

Linganisha coefficients za $\mathbf{q}$:

$$\frac{2h}{3} = 1 - \frac{7k}{12} \quad (2)$$

Badilisha (1) kwenye (2):

$$\frac{2h}{3} = 1 - \frac{7(h/2)}{12}$$ $$\frac{2h}{3} = 1 - \frac{7h}{24}$$

Zidisha pande zote mbili kwa 24 ili kuondoa madhehebu:

$$24\left(\frac{2h}{3}\right) = 24(1) - 24\left(\frac{7h}{24}\right)$$ $$16h = 24 - 7h$$ $$16h + 7h = 24$$ $$23h = 24$$ $$h = \frac{24}{23}$$

Sasa pata $k$ kwa kutumia $k = \frac{h}{2}$:

$$k = \frac{1}{2} \times \frac{24}{23} = \frac{12}{23}$$ $$\boxed{h = \frac{24}{23}, k = \frac{12}{23}}$$

e) Bainisha uwiano ambao X inagawanya QR.
Kutoka sehemu (c), tunajua $\vec{QX} = k\vec{QR}$.
Hii inamaanisha X inagawanya QR kwa uwiano $QX : XR$.
$\vec{QX} = k\vec{QR}$.
$\vec{XR} = \vec{QR} - \vec{QX} = \vec{QR} - k\vec{QR} = (1-k)\vec{QR}$.
Hivyo uwiano $QX : XR = k : (1-k)$.
Tunajua $k = \frac{12}{23}$.

$$1-k = 1 - \frac{12}{23} = \frac{23 - 12}{23} = \frac{11}{23}$$

Uwiano ni $\frac{12}{23} : \frac{11}{23}$.
Kuzidisha kwa 23, tunapata uwiano:

$$12 : 11$$ $$\boxed{\text{Uwiano ni } 12 : 11}$$