antwoorden

Onderstaand overzicht volgt de nummering en de opgaven van de derde editie.
[antwoorden eerste editie | antwoorden tweede editie]

1

2

We geven enkel de coördinaten van de vectoren:

a. $\vec a + \vec b = (0,3)$ c. $-\vec c + \tfrac{5}{2}\vec a = (6,\tfrac{11}{2})$
b. $2\vec c -1.5\vec b = (1,-9)$ d. $2(\vec a - 3 \vec b) - 2 \vec c = (18,-4)$


3
parametervoorstelling cartesische vergelijking
a. $\displaystyle \left\{ \begin{array}{l} x = 7\mu \\ y = -4\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $y = -\tfrac{4}{7}x$
b. $\displaystyle \left\{ \begin{array}{l} x = 1-3\mu \\ y = 3+2\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $y = -\tfrac{2}{3}x+\tfrac{11}{3}$
c. $\displaystyle \left\{ \begin{array}{l} x = 2+\mu \\ y = -2-8\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $y = -8x+14$
d. $\displaystyle \left\{ \begin{array}{l} x = -2 \\ y = 3-2\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $x = -2$
e. $\displaystyle \left\{ \begin{array}{l} x = 3+2\mu \\ y = 1-\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $y = -\tfrac{1}{2}x+\tfrac{5}{2}$
f. $\displaystyle \left\{ \begin{array}{l} x = -2 \\ y = 9\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $x=-2$
g. $\displaystyle \left\{ \begin{array}{l} x = -5 \\ y = 1+\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $x=-5$
h. $\displaystyle \left\{ \begin{array}{l} x = 4+2\mu \\ y = 2+3\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $y=\tfrac{3}{2}x-4$
i. $\displaystyle \left\{ \begin{array}{l} x = 3+\mu \\ y = 1+\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $y = x-2$
j. $\displaystyle \left\{ \begin{array}{l} x = 4+\mu \\ y = 1-\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $y = -x+5$
k. $\displaystyle \left\{ \begin{array}{l} x = 0 \\ y = \mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $x=0$
l. $\displaystyle \left\{ \begin{array}{l} x = -7+\mu \\ y = 2+\mu \end{array} \right. \quad (\mu \in \mathbb{R})$ $y = x+9$


4
a. $y = -\tfrac{1}{3}x+\tfrac{5}{3}$ b. $y = -x+2$ c. $x=0$


5
a. $p=7$ b. $p=1$ c. $p=3$


6
a. $5$ c. $2\sqrt{5}$ e. $2\sqrt{26}$ g. $\sqrt{157}$
b. $3\sqrt{29}$ d. $\sqrt{445}$ f. $2\sqrt{5}$ h. $2\sqrt{65}$


7

aanpassing opgave (eerste druk): evenwijdige rechten uit opgave 3h en 3i.

afstand tussen rechten uit 3h: $\frac{12}{\sqrt{13}}$
afstand tussen rechten uit 3i: $\frac{3\sqrt{2}}{2}$


8
a. $\displaystyle \frac{4\sqrt{5}}{5}$ b. $\displaystyle \frac{4\sqrt{5}}{5}$ c. $\displaystyle \sqrt{10}$


10
a. orthogonaal ($\vec u \cdot \vec v = 0$) b. niet orthogonaal ($\vec u \cdot \vec v = -11$) c. orthogonaal ($\vec u \cdot \vec v = 0$)


11
De scherpe hoek bedraagt $60^{\circ}$ of $\displaystyle \frac{\pi}{3}$ rad.


12
a. $\displaystyle (x-2)^2+(y+1)^2=9$ c. $\displaystyle y=-(x+2)^2+4$
b. $\displaystyle \frac{(x-3)^2}{4}+\frac{(y+3)^2}{5}=1$ d. $\displaystyle \frac{x^2}{9}-\frac{(y-2)^2}{4}=1$


13

Om de kegelsneden te interpreteren, herleiden we de vergelijking naar de standaardvorm.

a. $\displaystyle (x-2)^2+\left( y+\tfrac{1}{2} \right)^2=4$ d. $\displaystyle \frac{\left(x-\tfrac{1}{4}\right)^2}{\tfrac{1}{4}}-\frac{\left(y-\tfrac{1}{3}\right)^2}{\tfrac{1}{9}}=1$
b. $\displaystyle y=-3(x+4)^2-3$ e. $\displaystyle \frac{(x+3)^2}{9}-\frac{(y+1)^2}{4}=1$
c. $\displaystyle \frac{x^2}{16}+\frac{(y-2)^2}{4}=1$ f. $\displaystyle y=4\left( x-\tfrac{3}{2} \right)^2+\frac{1}{2}$