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In the figure below OA=a, OB=b, and BX meets OA or at C and BY:YA = 1:3. a) Express in terms of a and b: (1) BA (2) OY (3) BX. b) Given that OC = hOx and BC = kBX, determine the values of h and k.

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2 parts
a

a−b\mathbf{a} - \mathbf{b}

14a+34b\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}

12a−b\frac{1}{2}\mathbf{a} - \mathbf{b}

b

25\frac{2}{5}

45\frac{4}{5}

Step-by-step solution

7 steps

Haya, twende tukasuluhishe swali hili la vekta.

Taarifa tulizopewa:

  • OA⃗=a\vec{OA} = \mathbf{a}
  • OB⃗=b\vec{OB} = \mathbf{b}
  • XX iko kwenye OAOA na OX⃗=12OA⃗=12a\vec{OX} = \frac{1}{2}\vec{OA} = \frac{1}{2}\mathbf{a}.
  • YY iko kwenye ABAB na BY:YA=1:3BY:YA = 1:3.
  • Mstari BXBX na OYOY hukutana kwenye CC.

Part a· 3 steps

Eleza kwa kutumia a\mathbf{a} na b\mathbf{b}

  1. BA⃗\vec{BA}

  1. Tumia sheria ya pembetatu ya vekta.

    BA⃗=BO⃗+OA⃗\vec{BA} = \vec{BO} + \vec{OA} BA⃗=−OB⃗+OA⃗\vec{BA} = -\vec{OB} + \vec{OA} BA⃗=−b+a\vec{BA} = -\mathbf{b} + \mathbf{a} BA⃗=a−b\vec{BA} = \boxed{\mathbf{a} - \mathbf{b}}
    1. OY⃗\vec{OY}

  2. Tumia fomula ya sehemu ya vekta kwa pointi YY inayogawanya BABA kwa uwiano 1:31:3.

    OY⃗=3OB⃗+1OA⃗1+3\vec{OY} = \frac{3\vec{OB} + 1\vec{OA}}{1+3} OY⃗=3b+a4\vec{OY} = \frac{3\mathbf{b} + \mathbf{a}}{4} OY⃗=14a+34b\vec{OY} = \boxed{\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}}
    1. BX⃗\vec{BX}

  3. Tumia sheria ya pembetatu ya vekta.

    BX⃗=BO⃗+OX⃗\vec{BX} = \vec{BO} + \vec{OX} BX⃗=−OB⃗+12OA⃗\vec{BX} = -\vec{OB} + \frac{1}{2}\vec{OA} BX⃗=−b+12a\vec{BX} = -\mathbf{b} + \frac{1}{2}\mathbf{a} BX⃗=12a−b\vec{BX} = \boxed{\frac{1}{2}\mathbf{a} - \mathbf{b}}

Part b

25\frac{2}{5}

Kutokana na OC⃗=hOY⃗\vec{OC} = h\vec{OY} na BC⃗=kBX⃗\vec{BC} = k\vec{BX}, tafuta thamani za hh na kk.

  1. Eleza OC⃗\vec{OC} kwa njia mbili tofauti.

    Kutoka OC⃗=hOY⃗\vec{OC} = h\vec{OY}:

    OC⃗=h(14a+34b)\vec{OC} = h\left(\frac{1}{4}\mathbf{a} + \frac{3}{4}\mathbf{b}\right) OC⃗=h4a+3h4b(Equation 1)\vec{OC} = \frac{h}{4}\mathbf{a} + \frac{3h}{4}\mathbf{b} \quad \text{(Equation 1)}

    Kutoka BC⃗=kBX⃗\vec{BC} = k\vec{BX}, tunaweza kuandika OC⃗\vec{OC} kama:

    OC⃗=OB⃗+BC⃗\vec{OC} = \vec{OB} + \vec{BC} OC⃗=OB⃗+kBX⃗\vec{OC} = \vec{OB} + k\vec{BX} OC⃗=b+k(12a−b)\vec{OC} = \mathbf{b} + k\left(\frac{1}{2}\mathbf{a} - \mathbf{b}\right) OC⃗=b+k2a−kb\vec{OC} = \mathbf{b} + \frac{k}{2}\mathbf{a} - k\mathbf{b} OC⃗=k2a+(1−k)b(Equation 2)\vec{OC} = \frac{k}{2}\mathbf{a} + (1-k)\mathbf{b} \quad \text{(Equation 2)}

  2. Linganisha vigawo vya a\mathbf{a} na b\mathbf{b} kutoka Equation 1 na Equation 2.

    Kulinganisha vigawo vya a\mathbf{a}:

    h4=k2\frac{h}{4} = \frac{k}{2} h=2k(Equation 3)h = 2k \quad \text{(Equation 3)}

    Kulinganisha vigawo vya b\mathbf{b}:

    3h4=1−k(Equation 4)\frac{3h}{4} = 1-k \quad \text{(Equation 4)}

  3. Tatua milinganyo ya wakati mmoja kupata hh na kk.

    Badilisha Equation 3 ndani ya Equation 4:

    3(2k)4=1−k\frac{3(2k)}{4} = 1-k 6k4=1−k\frac{6k}{4} = 1-k 3k2=1−k\frac{3k}{2} = 1-k

    Kuzidisha pande zote mbili kwa 2:

    3k=2(1−k)3k = 2(1-k) 3k=2−2k3k = 2 - 2k 3k+2k=23k + 2k = 2 5k=25k = 2 k=25k = \boxed{\frac{2}{5}}

  4. Pata thamani ya hh kwa kutumia Equation 3.

    h=2kh = 2k h=2(25)h = 2\left(\frac{2}{5}\right) h=45h = \boxed{\frac{4}{5}}
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