1. ## subspace 2

Show that the set of all points in the plan ax + by + cz = 0 is a subspace of R^3. Find vector v1 and v2 such that [v1, v2] is the plan 2x - 3y + 4z = 0

2. See if this resolution is correct?

$\displaystyle u=(a1,b1,c1)$
$\displaystyle v=(a2,b2,c2)$

$\displaystyle u+v=(a1+a2,b1+b2,c1+c2)$ belongs to the subspace of R^3

$\displaystyle \alpha . u=( \alpha a1, \alpha b1, \alpha c1)$ belongs to the subspace of R^3

$\displaystyle v1=(1,-\frac{3}{2},2)$
$\displaystyle v2=(1,-\frac{3}{2},2)$

$\displaystyle v1+v2=(2,-3,4)$

$\displaystyle \alpha . u = (2,-3,4)$

Is correct?

3. Originally Posted by Apprentice123
See if this resolution is correct?

$\displaystyle u=(a1,b1,c1)$
$\displaystyle v=(a2,b2,c2)$

$\displaystyle u+v=(a1+a2,b1+b2,c1+c2)$ belongs to the subspace of R^3

$\displaystyle \alpha . u=( \alpha a1, \alpha b1, \alpha c1)$ [COLOR=Red]belongs to the subspace of R^3
The solution looks correct but you need to clarify some things, firstly your two vectors should be any two vectors I think you may have implied this but i am confused by your notation here. Also you haven't given a reason using the vector space properties or the equation of the plane for why the sum belongs to the plane. You have also forgot to make any mention of the zero vector.

Bobak

4. How can I find the vector that is of the plan

5. Originally Posted by Apprentice123
How can I find the vector that is of the plan
Well suppose you just have two vectors on the plane. $\displaystyle u =( x_1 , y_1 , z_1)$ and $\displaystyle v =( x_2 , y_2 , z_2)$. them being on the plane means that $\displaystyle a x_i + b y_i + c z_i = 0$.

so$\displaystyle u + v = ( x_1 + x_2 , y_1 + y_2 , z_1 + z_2 )$ now you must show that this is on the plane.
$\displaystyle a (x_1 + x_2 ) + b(y_1 + y_2) + c (z_1 + z_2)$
$\displaystyle = a x_1 + b y_1 + c z_1 + a x_2 + b y_2 + c z_2 = 0 + 0 = 0$

Therefore the sum of any two vectors on the plane is also on the plane.

Bobak

6. then I can have:

$\displaystyle v1=v2=(1,-\frac{3}{2},2)$ ?

7. Originally Posted by Apprentice123
then I can have:

$\displaystyle v1=v2=(1,-\frac{3}{2},2)$ ?
No, neither v1 or v2 are in the plane, unless I have misunderstood the question.

Bobak

8. How can I determine v1 and v2 so that they are in the plan?