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Let $\alpha,\beta,a,b,c \in \mathbb{R}.$ Consider three affine planes in the affine space $\mathbb{R}^{3}$: $P_{1}$ of equation $x+2y+\beta z=a$, $P_2$ of equation $2x+y=b$, $P_{3}$ of equation $\alpha x +(\alpha+1)y=c.$

What is the dimension of their intersection $F=P_1 \cap P_2 \cap P_3$? When F is not empty, give it as a system of equations (with the fewest possible number of equations).

my attemps

to calculate dimensional of $F$ i thing we have show that $P_1,P_2,P_3$ are affine subspace of $R^{3}$ independantes Then we can write $dim F=dim(P_1∩P_2∩P_3)=$

To show that have show that $P_1,P_2,P_3$ are affine subspace of $R^{3}$ independent we hae to show that $P_1=kerf_{1}$ with $f_1$ is a linear form non-zero and $P_2=kerf_{2}$ with $f_2$ is a linear form non-zero and $P_3=kerf_{3}$ with $f_3$ is a linear form non-zero

how we can show tha f_1 f_2 f_3 are lineare forme non zero independent

for example we can consider $P_1=Ker f_1$ with

$$\begin{array}{ccccc} f_1 & : & R^3 & \longrightarrow & R \\ & & (x,y,z) & \mapsto & f(x,y,z)=x+2y+\beta. z=a \\ \end{array}$$

we have to show $f_1$ is linear forme non zero

if $a=$ is vector space

if possible someone suggest detailed answer

any help would be appreciated!

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  • $\begingroup$ The intersection is the set of points $(x,y,z)$ satisfying $$\begin{pmatrix} 1&2&\beta \\ 2&1&0 \\ \alpha&\alpha+1&0\end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} a\\b\\c \end{pmatrix}.$$How do you find the solution set to such an equation? $\endgroup$ – Greg Martin Dec 10 '14 at 20:40
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    $\begingroup$ i thing the easy one is Gauss–Seidel method $\endgroup$ – Educ Dec 10 '14 at 20:56
  • $\begingroup$ @GregMartin how can i explain to my friend that The intersection ? is the set of points (x,y,z) satisfying that system can you elaborate this please $\endgroup$ – Educ Dec 11 '14 at 20:41
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Un point de coordonnées $(x,y,z)$ appartient à l'intersection des trois plans s'il vérifie les trois équations, c'est-à-dire le système proposé par Greg Martin. Ce système se résout facilement par la méthode de Gauss. Selon les valeurs des paramètres alpha et bêta le rang $r$ du système est différent et la dimension de $F$ est $3 -r$. Voyez http://wims.auto.u-psud.fr/wims/wims.cgi?module=U1/algebra/docsyslin.fr

L'enseignante auteur de l'exercice


A point with coordinates $(x,y,z)$ belongs to the intersection of the three planes if it satisfies the three equations, that is, the system

$$\begin{pmatrix} 1&2&\beta \\ 2&1&0 \\ \alpha&\alpha+1&0\end{pmatrix} \begin{pmatrix} x\\y\\z \end{pmatrix} = \begin{pmatrix} a\\b\\c \end{pmatrix}$$

proposed by Greg Martin. That system is easily solved by Gauß elimination. The rank $r$ of the system depends on the value of the parameters $\alpha$ and $\beta$ and the dimension of $F$ is $3-r$. See http://wims.auto.u-psud.fr/wims/wims.cgi?module=U1/algebra/docsyslin.fr

The teacher and author of the exercise.

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