# Thread: [SOLVED] I think I need some serious help...lol

1. ## [SOLVED] I think I need some serious help...lol

Textbook----------

x + 3y + 5z = 4
x + 2y - 3z = 5
2x + 5y + 2z =8

This system has no solution.
The planes in the row picture don't meet at a point.
End of Textbook------------

Ok.

Three vectors fill a 3D space if all three at independent, ie they are not linear combinations of each other.

Three vectors fill a plane if 2 are independent and 1 is a linear combination of the other two.

Take equation one.
x + 3y + 5z = 4
which is the same as
x[1] + y[3] + z[5] = [4]

I think I may be retarded here...

[3] is a linear combination of [1] of the form 3[1]
Therefore x[1] + y[3] are dependent vectors. *** have a feeling this might be wrong.
[5] is a linear combination of 2[1] and 3[1]
Therefore x[1] + y[3] + z[5] are all dependent vectors...

So how the heck do they make a plane? Wouldn't they just make a line?
If all three equations make lines, and the lines don't intersect, I understand. But the thing that's throwing me off is from the text: "The planes in the row picture don't meet at a point."

Is there something I'm not getting that is blatantly obvious?

2. Originally Posted by Noxide
Textbook----------

x + 3y + 5z = 4
x + 2y - 3z = 5
2x + 5y + 2z =8

This system has no solution.
The planes in the row picture don't meet at a point.
End of Textbook------------

Ok.

Three vectors fill a 3D space if all three at independent, ie they are not linear combinations of each other.

Three vectors fill a plane if 2 are independent and 1 is a linear combination of the other two.

Take equation one.
x + 3y + 5z = 4
which is the same as
x[1] + y[3] + z[5] = [4]

I think I may be retarded here...

[3] is a linear combination of [1] of the form 3[1]
Therefore x[1] + y[3] are dependent vectors. *** have a feeling this might be wrong.
[5] is a linear combination of 2[1] and 3[1]
Therefore x[1] + y[3] + z[5] are all dependent vectors...

So how the heck do they make a plane? Wouldn't they just make a line?
If all three equations make lines, and the lines don't intersect, I understand. But the thing that's throwing me off is from the text: "The planes in the row picture don't meet at a point."

Is there something I'm not getting that is blatantly obvious?
x + 3y + 5z = 4 is NOT a vector.

"Therefore x[1] + y[3] + z[5] are all dependent vectors..."

No they are not. The combination x[1] + y[3] + z[5] can be interpreted as a SINGLE vector whose components when resolved in the x, y, z directions as 1, 3, 5. Hence it can be written in row form as (1 3 5).

There are two other vectors here: (1 2 -3) and (2 5 2).

The vector (2 5 2) is the sum of (1 3 5) and (1 2 -3), or, if you like, (1 3 5) + (1 2 -3) - (2 5 2) = 0.

So a linear combination of these vectors adds up to zero and therefore the vectors are dependent.

3. Thanks!
I'm just going to run by you what I have learned from your post. Let me know if I am correct.

x[1]+ y[3] + z[5] can be interpreted as a single vector
The single vector is [4]

x + 3y + 5z=4 is not a vector
It is a linear combination of column vectors resulting in a vector, [4]

components of an n-dimensional vector when resolved in n directions can be written in row form (n1, n2, n3, ... , nn)

[a]
[b] written in row form is (a, b, c)
[c]

I'm a bit fuzzy after this line ____________

when a linear combination of vectors adds to zero, the vectors are dependent

and since 9 =/= 8 there is no solution to this sytem?

4. Originally Posted by Noxide
Thanks!
I'm just going to run by you what I have learned from your post. Let me know if I am correct.

x[1]+ y[3] + z[5] can be interpreted as a single vector
The single vector is [4]

x + 3y + 5z=4 is not a vector
It is a linear combination of column vectors resulting in a vector, [4]

components of an n-dimensional vector when resolved in n directions can be written in row form (n1, n2, n3, ... , nn)

[a]
[b] written in row form is (a, b, c)
[c]

I'm a bit fuzzy after this line ____________

when a linear combination of vectors adds to zero, the vectors are dependent

and since 9 =/= 8 there is no solution to this sytem?
"x[1]+ y[3] + z[5] can be interpreted as a single vector
The single vector is [4]"

No the single vector is NOT 4.

x[1]+ y[3] + z[5] (using the notation you chose for it) is a vector.

x + 3y + 5z=4 IS NOT A VECTOR!!!

It's the equation of a plane.

You're trying to do two different things here:

1. Find the point of intersection (if there is one) of three planes, as was given in the initial question. These are NOT vectors.

2. Determine whether the vectors whose components are the same as those for the equations for the planes are dependent or independent.

The two are in a sense analogous. The planes have a point of intersection if and only if the vectors are independent. They're not so there isn't.

Whether those 3 planes intersect in a line is something else that needs to be worked out, which you did by determining whether there is a solution to the equations, and there isn't so they don't.

5. That was rather embarassing!