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Re: Re: Re: Re: Gedeelte van examenvraag mbo 78-79 (2)

Dit is een reactie op vraag 89588

Is goed uiteindelijk dus (2¹/₂,5,3)+m (2,2,1). Het model geeft (-3¹/₂,-1,0)+m(2,2,1) met de toevoeging bijvoorbeeld...
mboudd
Leerling mbo - zaterdag 11 april 2020

Antwoord

$
\begin{array}{l}
P(\lambda + \mu ,2\lambda ,3 + \mu ) \\
A(1,0,1) \\
B(0,2, - 1) \\
d(A,P) = d(B,P) \\
d(A,P) = \sqrt {\left( {\lambda + \mu - 1} \right)^2 + \left( {2\lambda } \right)^2 + \left( {3 + \mu - 1} \right)^2 } \\
d(A,P) = \sqrt {5\lambda ^2 + 2\lambda \mu - 2\lambda + 2\mu ^2 + 2\mu + 5} \\
d(B,P) = \sqrt {\left( {\lambda + \mu } \right)^2 + \left( {2\lambda - 2} \right)^2 + \left( {3 + \mu + 1} \right)^2 } \\
d(B,P) = \sqrt {5\lambda ^2 + 2\lambda \mu - 8\lambda + 2\mu ^2 + 8\mu + 20} \\
\sqrt {5\lambda ^2 + 2\lambda \mu - 2\lambda + 2\mu ^2 + 2\mu + 5} = \sqrt {5\lambda ^2 + 2\lambda \mu - 8\lambda + 2\mu ^2 + 8\mu + 20} \\
5\lambda ^2 + 2\lambda \mu - 2\lambda + 2\mu ^2 + 2\mu + 5 = 5\lambda ^2 + 2\lambda \mu - 8\lambda + 2\mu ^2 + 8\mu + 20 \\
- 2\lambda + 2\mu + 5 = - 8\lambda + 8\mu + 20 \\
6\lambda = 6\mu + 15 \\
\lambda = \mu + 2\frac{1}{2} \\
l:\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
0 \\
0 \\
3 \\
\end{array}} \right) + \left( {\mu + 2\frac{1}{2}} \right)\left( {\begin{array}{*{20}c}
1 \\
2 \\
0 \\
\end{array}} \right) + \mu \left( {\begin{array}{*{20}c}
1 \\
0 \\
1 \\
\end{array}} \right) \\
l:\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
0 \\
0 \\
3 \\
\end{array}} \right) + 2\frac{1}{2}\left( {\begin{array}{*{20}c}
1 \\
2 \\
0 \\
\end{array}} \right) + \mu \left( {\begin{array}{*{20}c}
1 \\
2 \\
0 \\
\end{array}} \right) + \mu \left( {\begin{array}{*{20}c}
1 \\
0 \\
1 \\
\end{array}} \right) \\
l:\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
0 \\
0 \\
3 \\
\end{array}} \right) + \left( {\begin{array}{*{20}c}
{2\frac{1}{2}} \\
5 \\
0 \\
\end{array}} \right) + \mu \left( {\begin{array}{*{20}c}
2 \\
2 \\
1 \\
\end{array}} \right) \\
l:\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{2\frac{1}{2}} \\
5 \\
3 \\
\end{array}} \right) + \mu \left( {\begin{array}{*{20}c}
2 \\
2 \\
1 \\
\end{array}} \right) \\
\end{array}
$

Kijk 's aan...

Of op de andere manier:

$
\begin{array}{l}
AB:\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
1 \\
0 \\
1 \\
\end{array}} \right) - \left( {\begin{array}{*{20}c}
0 \\
2 \\
{ - 1} \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
1 \\
{ - 2} \\
2 \\
\end{array}} \right) \\
\alpha :\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{\frac{1}{2}} \\
1 \\
0 \\
\end{array}} \right) + \rho \left( {\begin{array}{*{20}c}
2 \\
1 \\
0 \\
\end{array}} \right) + \tau \left( {\begin{array}{*{20}c}
2 \\
0 \\
{ - 1} \\
\end{array}} \right) \\
\beta :\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
0 \\
0 \\
3 \\
\end{array}} \right) + \lambda \left( {\begin{array}{*{20}c}
1 \\
2 \\
0 \\
\end{array}} \right) + \mu \left( {\begin{array}{*{20}c}
1 \\
0 \\
1 \\
\end{array}} \right) \\
\left\{ \begin{array}{l}
\frac{1}{2} + 2\rho + 2\tau = \lambda + \mu \\
1 + \rho = 2\lambda \\
- \tau = 3 + \mu \\
\end{array} \right. \\
\left\{ \begin{array}{l}
1 + 4\rho + 4\tau = 2\lambda + 2\mu \\
1 + \rho = 2\lambda \\
- 2\tau = 6 + 2\mu \\
\end{array} \right. \\
(1) - (3) \\
1 + 4\rho + 6\tau = 2\lambda - 6 \\
7 + 4\rho + 6\tau = 2\lambda \\
(2) \\
1 + \rho = 7 + 4\rho + 6\tau \\
- 3\rho = 6 + 6\tau \\
\rho = - \frac{{6 + 6\tau }}{3} = - 2 - 2\tau \\
\alpha :\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{\frac{1}{2}} \\
1 \\
0 \\
\end{array}} \right) + \left( { - 2 - 2\tau } \right)\left( {\begin{array}{*{20}c}
2 \\
1 \\
0 \\
\end{array}} \right) + \tau \left( {\begin{array}{*{20}c}
2 \\
0 \\
{ - 1} \\
\end{array}} \right) \\
\alpha :\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{\frac{1}{2}} \\
1 \\
0 \\
\end{array}} \right) + - 2\left( {\begin{array}{*{20}c}
2 \\
1 \\
0 \\
\end{array}} \right) - 2\tau \left( {\begin{array}{*{20}c}
2 \\
1 \\
0 \\
\end{array}} \right) + \tau \left( {\begin{array}{*{20}c}
2 \\
0 \\
{ - 1} \\
\end{array}} \right) \\
\alpha :\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{ - 3\frac{1}{2}} \\
{ - 1} \\
0 \\
\end{array}} \right) + \tau \left( {\begin{array}{*{20}c}
{ - 4} \\
{ - 2} \\
0 \\
\end{array}} \right) + \tau \left( {\begin{array}{*{20}c}
2 \\
0 \\
{ - 1} \\
\end{array}} \right) \\
\alpha :\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{ - 3\frac{1}{2}} \\
{ - 1} \\
0 \\
\end{array}} \right) + \tau \left( {\begin{array}{*{20}c}
{ - 2} \\
{ - 2} \\
{ - 1} \\
\end{array}} \right) \\
\alpha :\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{ - 3\frac{1}{2}} \\
{ - 1} \\
0 \\
\end{array}} \right) + \tau \left( {\begin{array}{*{20}c}
2 \\
2 \\
1 \\
\end{array}} \right) \\
\end{array}
$

't Is hard werken. Hopelijk helpt het...

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zaterdag 11 april 2020