Klart det.
Vi skal ha mer omfattende prøve for kap 5+6 på tirsdag, men her er løsningsforslag for denne:
Oppgave 1
a) Den får du klare selv.
b)
[tex]\frac12\left(\vec a - \frac32 \vec b\right) - 2\left(\frac14\vec a - \frac12\vec b\right) \\ = \ \frac13\vec a - \frac12 \vec b - \frac12 \vec a + \vec b \\ = \ \left(\frac26 - \frac36\right)\vec a + \frac12 \vec b = \underline{\underline{-\frac16\vec a + \frac12 \vec b}}[/tex]
c)
[tex]\vec{AB} = \vec a \ \ \ \ \vec{AC}=\vec b[/tex]
(1)
[tex]\vec{AD} = \vec{AB} + \frac14\vec{BC} = \vec a + \frac14\left(-\vec a + \vec b\right) \\ \underline{\underline{= \frac34\vec a + \frac14 \vec b}}[/tex]
(2)
[tex]\vec{AE} = \frac1{10}\vec b \ \ \ \ \vec{BF} = -\frac12\vec{BC} \\ \vec{EF} = \vec{EA} + \vec{AB} + \vec{BF} = -\frac{1}{10}\vec b + \vec a - \frac12\vec{BC} \\ = -\frac{1}{10}\vec b + \vec a - \frac12\left(-\vec a + \vec b\right) = -\frac{1}{10}\vec b + \vec a + \frac12\vec a - \frac12\vec b \\ \underline{\underline{= \frac32\vec a - \frac35\vec b}}[/tex]
Oppgave 2
[tex]A(2, -1) \ \ \ B(4,1) \ \ \ C(3,3)[/tex]
a)
[tex]\vec{AB} = \left[4-2, 1-(-1)\right] = \underline{\underline{\left[2, 2\right]}} \\ \vec{BC} = \left[3-4, 3-1\right] = \underline{\underline{\left[-1, 2\right]}} \\ \vec{AC} = \left[3-2, 3-(-1)\right] = \underline{\underline{\left[1, 4\right]}}[/tex]
b)
[tex]\left|\vec{AB}\right| = \sqrt{2^2 + 2^2} = \sqrt{8} = \sqrt{4\cdot 2} = \underline{\underline{2\sqrt{2}}} \\ \left|\vec{BC}\right| = \sqrt{(-1)^2 + 2^2} = \underline{\underline{\sqrt{5}}}[/tex]
c)
M = midtpunktet på AC
[tex]\vec{OM} = \vec{DA} + \frac12\vec{AC} = \left[2, -1\right] + \frac12\left[1, 4\right] = \left[2+\frac12, -1+\frac42\right] = \left[\frac52, 1\right] \\ \underline{\underline{\Rightarrow M\left(\frac52, 1\right)}}[/tex]
d)
[tex]D(1, y) \ \ \text{avstand} \ \sqrt{2} \ \text{til A}[/tex]
[tex]\vec{AD} = \left[1-2, y-(-1)\right] = \left[-1, y+1\right] \\ \left|\vec{AD}\right| = \sqrt{(-1)^2 + (y+1)^2} = \sqrt{1^2 + y^2 + 2y + 1} = \sqrt{y^2 + 2y + 2} = \sqrt{2}[/tex]
(fordi [tex]\small{|\vec{AB}| = \sqrt{2}}[/tex])
[tex]\Rightarrow y^2 + 2y + 2 = 2 \\ y^2 + 2y = 0 \\ y(y+2) = 0 \\ \Rightarrow \underline{\underline{y=0}} \ \text{eller} \ \underline{\underline{y=-2}}[/tex]
Oppgave 3
[tex]\vec a = \left[-1, 2\right] \ \ \ \ \vec b = \left[3, -4\right][/tex]
[tex]\vec a - \vec b = \left[-1, 2\right]-\left[3, -4\right] = \left[-1-3, 2-(-4)\right] = \underline{\underline{\left[-4, 6\right]}}[/tex]
