# Linear Independence: Is this true?

1. Oct 17, 2012

### Bipolarity

Consider a plane P in $ℝ^{3}$. Is it necessarily the case that any vector outside this plane cannot be expressed as a linear combination of finitely many vectors on this plane?

I would think yes; if you tried to parametrize the plane P with two parameters, could we somehow show that there are no values of the parameters for which the linear combination of the vectors on the plane equals the vector outside the plane?

But I need an expert's opinion on this.

BiP

2. Oct 17, 2012

### Stephen Tashi

What do you mean by "outside" the plane? Are you visualizing a plane in $R^3$ that does not pass through the origin and visualizing a vector "in the plane" as one that is represented by a directed line segment that is in the plane? Or are you defining a vector "in the plane" to mean a vector from the origin to a point on the plane? - when the equation of a plane is given in terms of vectors, it is usually done by writing a relation that is satisified by vectors whose "tail" is the origin and whose "head" is a point on the plane.

3. Oct 17, 2012

### Bipolarity

I am actually referring to the former: "directed line segment in the plane". That is, if you take two points in the plane and take their "difference", then you get a vector which is "in the plane". I thought this was standard used in constructing parametrizations of planes in 3-space. Or perhaps I am not very clear in my conceptualization of vectors.

BiP

4. Oct 17, 2012

### Stephen Tashi

You are visualizing vectors as "free vectors". I've never seen a systematic elementary exposition of how to deal with them. Illustrations in physics books sometimes draw vectors so their "tail" doesn't begin at the origin of the coordinate system. That type of vectoris called a "free vector". All the common formula involving cartesian coordinates of vectors assume that the "tail" is at the origin. This is the standard way of treating vectors. They are "bound" vectors.

If you regard the vector between the points (3,5) and (4,9) as (1,4) and you regard (1,4) as referring to vector whose tail is at (0,0) and whose head is at (1,4) then (1,4) does not lie on the line segment connecting (3,5) to (4,9). In fact, if you are only using 2 numbers to represent a 2D vector, then you don't have enough numbers to represent both the head and tail of the vector.

Likewise, if you are using 3 numbers to represent a vector in $R^3$ you can't represent both the tail and the head. When we say "a vector in $R^3$" , this is usually taken to mean "a bound vector specified by 3 numbers insead of a free vector specified by 6 numbers.

In terms of bound vectors, a plane in $R^3$ not passing through the origin can be represented by taking all the vectors that lie in a given plane that does pass through the origin and adding some constant vector $V_c$ to them.

When you ask about a linear combination of vectors , you have to specificy what vectors we can use in the linear combination. A linear combination of "bound" vectors is always a vector whose tail is at the origin. Multiples of a "bound" vector never lie "in" a plane that doesn't pass through the origin..