This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Huu ni swali la vekta. Tutatumia taarifa zilizotolewa na sheria za vekta kutatua kila sehemu. Taarifa Zilizotolewa: Trapezium PQRS. PQ = 3ST. Hii inamaanisha PQ ni sambamba na ST na urefu wake ni mara 3 ya ST. Kwa kuwa T iko kwenye SR, hii inamaanisha PQ ni sambamba na SR. T inagawanya SR kwa uwiano wa 4:1. Hivyo ST = (4)/(5)SR na TR = (1)/(5)SR. U ni katikati ya QT. Hivyo QU = UT = (1)/(2)QT. PU na QR zinakutana kwenye X. Hivyo X iko kwenye PU na X iko kwenye QR. PX = hPU na QX = kQR. PQ = q na PS = p. Hatua za Kwanza: Kutoka PQ = 3ST: ST = (1)/(3)PQ = (1)/(3)q Kutoka T inagawanya SR kwa uwiano 4:1, na ST = (4)/(5)SR: SR = (5)/(4)ST = (5)/(4)((1)/(3)q) = (5)/(12)q a) Eleza QR kwa kutumia p na q. Tunatumia njia ya vekta kutoka Q hadi R: QR = QP + PS + SR Tunajua QP = -PQ = -q. QR = -q + p + (5)/(12)q QR = p - (7)/(12)q QR = p - (7)/(12)q b) Eleza PX kwa kutumia p, q na h. X iko kwenye PU, na PX = hPU. Kwanza tunahitaji kupata PU. PU = PS + SU Ili kupata SU, tunahitaji ST na TU. SU = ST + TU U ni katikati ya QT, hivyo TU = (1)/(2)TQ = -(1)/(2)QT. Kwanza, pata QT: QT = QP + PS + ST QT = -q + p + (1)/(3)q QT = p - (2)/(3)q Sasa, pata TU: TU = -(1)/(2)(p - (2)/(3)q) = -(1)/(2)p + (1)/(3)q Sasa, pata SU: SU = (1)/(3)q + (-(1)/(2)p + (1)/(3)q) = -(1)/(2)p + (2)/(3)q Sasa, pata PU: PU = p + (-(1)/(2)p + (2)/(3)q) = (1)/(2)p + (2)/(3)q Mwishowe, pata PX: PX = hPU = h((1)/(2)p + (2)/(3)q) PX = (h)/(2)p + (2h)/(3)q c) Eleza PX kwa kutumia p, q na k. X iko kwenye QR, na QX = kQR. Tunatumia njia ya vekta kutoka P hadi X: PX = PQ + QX PX = PQ + kQR Tunajua PQ = q na kutoka sehemu (a), QR = p - (7)/(12)q. PX = q + k(p - (7)/(12)q) PX = q + kp - (7k)/(12)q PX = kp + (1 - (7k)/(12))q PX = kp + (1 - (7k)/(12))q d) Pata thamani za h na k. Tunalinganisha milinganyo miwili ya PX kutoka sehemu (b) na (c): (h)/(2)p + (2h)/(3)q = kp + (1 - (7k)/(12))q Kwa kuwa p na q si vekta sambamba, coefficients zao lazima ziwe sawa. Linganisha coefficients za p: (h)/(2) = k (1) Linganisha coefficients za q: (2h)/(3) = 1 - (7k)/(12) (2) Badilisha (1) kwenye (2): (2h)/(3) = 1 - (7(h/2))/(12) (2h)/(3) = 1 - (7h)/(24) Zidisha pande zote mbili kwa 24 ili kuondoa madhehebu: 24((2h)/(3)) = 24(1) - 24((7h)/(24)) 16h = 24 - 7h 16h + 7h = 24 23h = 24 h = (24)/(23) Sasa pata k kwa kutumia k = (h)/(2): k = (1)/(2) × (24)/(23) = (12)/(23) h = (24)/(23), k = (12)/(23) e) Bainisha uwiano ambao X inagawanya QR. Kutoka sehemu (c), tunajua QX = kQR. Hii inamaanisha X inagawanya QR kwa uwiano QX : XR. QX = kQR. XR = QR - QX = QR - kQR = (1-k)QR. Hivyo uwiano QX : XR = k : (1-k). Tunajua k = (12)/(23). 1-k = 1 - (12)/(23) = (23 - 12)/(23) = (11)/(23) Uwiano ni (12)/(23) : (11)/(23). Kuzidisha kwa 23, tunapata uwiano: 12 : 11 Uwiano ni 12 : 11 Tuma swali linalofuata 📸