Question

Find two vectors that form a basis for the plain represented by the equation x + 2y + 3z = 0.

Answer 1

sp. plane NOT plain, doh, that's embarrassing!

Answer 2

you can take any two vectors lineary indepedant (except (000))...and "belong both to " the plane
for example...(0,-3,2)and (0,-2,3)
there are different ways to this question :) !

Answer 3

Neemo, that doesn't seem to be correct. No combination of those two vectors, (0,3,-2) and (0,-2,3), can give an x value other than zero. Yet, clearly in the equation x + 2y+ 3z = 0, x can have values other than zero.

Answer 4

yeaah sorry ; I made a mistake...the vectors (0,3;-2) and (0,-2,3) are not linearly independent...(0;3;-2)*2+3*(0,-2,3)=000
we can choose (0,3,-2) and (1,4,-3) for exampl !

Answer 5

I think the two vectors from your first answer are in fact linear independent.
Note: \[2\left[\begin{matrix}0 \\ 3 \\ -2\end{matrix}\right] + 3\left[\begin{matrix}0 \\ -2 \\ 3\end{matrix}\right] =\]
\[\left[\begin{matrix}0 \\ 6 \\ -4\end{matrix}\right] + \left[\begin{matrix}0 \\ -6 \\ 9\end{matrix}\right] =\]
\[\left[\begin{matrix}0 \\ 0 \\ 5\end{matrix}\right] != 0\]

Additionally elimination by row operations
\[\left[\begin{matrix}0 & 0 \\ 3 & -2 \\ -2 & 3\end{matrix}\right] => \left[\begin{matrix}0 & 0 \\ 3 & -2 \\ 0 & 5/3\end{matrix}\right]\]

demonstrates r = 2 and that the two vectors are linearly independent.

I think your original answer certainly was a plane in \[R^{3}\]... just not the same plane as described by the equation x + 2y + 3z = 0.

I believe your new answer is correct, but is there a systematic way to find the answer in general?

Answer 6

I was sleepy when I wrote my first answer....the vector(0,-2,3) doesn't belong to the plane...but if you have a plane...certainly , it's of dimension 2 , so , if you found two vectors linearly independent ( and also belong to the plaaaane !!!!) then , it's a basis of this plane...
in general; I don't knwo if you are familiar with this ( but the prooof is simple )
if dim(S)=n and you've found n nonzero vectors linearly independent,(or spans S) and belong to S ! Then a collection of these vectors is a basis of S !

Answer 7

Sorry to intrude here. x + 2y + 3z = 0 when written in matrix form would look like: \[\left[\begin{matrix}1 & 2 & 3 \\ \end{matrix}\right] \]
The rank of the above matrix is r = 1. So, the null-space is n-r = 3 - 1 = 2D space. To find the basis of the null-space of the above matrix, set first free-variable (second col) to 1 and second free-variable (third col) to 0. This gives the vector (-2,1,0). Similarly, setting second free-variable to 1 and first to 0, we get the vector (-3,0,1). Thus the vectors (-2,1,0) and (-3,0,1) are the two vectors that form a basis to the plane x + 2y + 3z = 0

Answer 8

Thank you for intruding! I asked this question because somehow I had lost the connection between the plot of a function with several symbols, and the linear algebra of matrices.

If I had written the equation x + 2y + 3z = 5, it would still be a plane in R^3, and setting the third variable the free variables to 1 and 0 would find two independent vectors in that plane (without finding the nullspace). Of course, we could subtract 5 from both sides to get x + 2y + 3z - 5 = 0... then could we find the null space for Ax = 0 with a modified x the same way?

\[\left[\begin{matrix}1 & 2 & 3 & 5\end{matrix}\right]\left[\begin{matrix}x \\ y \\ z \\ 1\end{matrix}\right] =[0]\]

Answer 9

Well, if the equation is x + 2y + 3z = 5, then we'd be trying to solve Ax = b. The general solution would be x_particular + x_nullspace. Just from calculating in my head (so not 100% sure), x_particular in this case would be (5,0,0) and to this you could add the null-space vectors that were determined previously (i.e (-2,1,0) and (-3,0,1)) which would result in the following basis:
basis 1: (3,1,0) and basis 2: (2,0,1)).

I am, however, not sure if you could put the 5 to LHS as you did (I notice also that you have missed a minus sign for the 1 in your column vector. It should be (x,y,z,-1)T, but that's trivial). I couldnt say if your approach is correct, sorry! Hope my reply helps!

Answer 10

I haven't read this whole thread yet, but I'll put in my 2 cents. We have
x + 2y + 3z = 0
which is the equation of a plain in R3 that goes through the origin and has the normal vector

N = <1,2,3>

I find vectors in the plane by guessing, since the dot product with N must be equal to 0. For example:

<1,1,-1>
<2,-1,0>
<1,-2,1>

Now, it seems clear by inspection that these vectors, taken two at a time, are independent, although since they all lie in a plane, while taken three at a time, they cannot be independent.

To check my work, I'll take two of them and compute a perpendicular vector using the trick of the "cross product" (it is not a real cross product but a way of remembering what to do):

\[\begin{bmatrix} i & j & k \\ 1 & 1 & -1 \\ 2 & -1 & 0 \end{bmatrix}\]

Compute the determinant. It is:
(0 - 1)i -(0 + 2)j +(-1 - 2)k
<-1,-2,-3>
This is (-1) times N, which is parallel to N and a perfectly good normal vector.