### Vector calculus in plane and space: Bases, Coordinates and Equations

### The notion of basis

In order to carry out actual calculations it is useful to describe vectors using numbers. The concepts we require for this are *basis* and *coordinate.*

Basis of plane and space

- A
**basis**(plural: bases) of the plane consists of two vectors #\vec{e}_1# and #\vec{e}_2#, which are not on the same line through #\vec{0}#. Each vector #\vec{v}# of the plane can now be written in its own unique way as a linear combination of the vectors #\vec{e}_1# , #\vec{e}_2#:

\[

\vec{v}= v_1 \cdot \vec{e}_1 + v_2 \cdot \vec{e}_2

\]

for certain numbers #v_1# and #v_2# unique to #\vec{v}#. The numbers #v_1# , #v_2# are called the**coordinates**of the vector #\vec{v}# relative to the basis. - A
**basis**of the space consists of three vectors #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3#, which are not in the same plane as #\vec{0}#. Each vector #\vec{v}# can be written as a linear combination of these three vectors:

\[

\vec{v} = v_1 \cdot\vec{e}_1 + v_2 \cdot\vec{e}_2 + v_3 \cdot\vec{e}_3

\]

where the scalars #v_1#, #v_2#, and #v_3# are uniquely determined. The vectors #\vec{e}_1#, #\vec{e}_2#, and #\vec{e}_3# form a basis of the space, and the numbers #v_1#, #v_2#, and #v_3#, are called the**coordinates**of the vector #\vec{v}# with respect to the basis. - The vectors #\vec{e}_1=\rv{1,0}# and #\vec{e}_2=\rv{0,1}# form the basis of #\mathbb{R}^2#, which we call the
**standard basis**of #\mathbb{R}^2#. - The vectors #\vec{e}_1=\rv{1,0,0}#, #\vec{e}_2=\rv{0,1,0}#, and #\vec{e}_3=\rv{0,0,1}#, form the basis of #\mathbb{R}^3#, which we call the
**standard basis**of #\mathbb{R}^3#.

The coordinates in the plane #\mathbb{R}^2# and in the space #\mathbb{R}^3#, which have been *previously* defined, are the coordinates as defined here for the standard basis:\[\rv{v_1,v_2}=v_1\cdot\rv{1,0}+v_2\cdot\rv{0,1}=v_1\cdot\vec{e}_1+v_2\cdot\vec{e}_2\] for #\mathbb{R}^2# and similarly, but with an additional term #v_3\cdot\vec{e}_3#, for #\rv{v_1,v_2,v_3}# in #\mathbb{R}^3#.

We will now check why the coordinates are uniquely determined by a basis of the plane: say #v_1\cdot \vec{e}_1+v_2\cdot\vec{e}_2=w_1\cdot\vec{e}_1+w_2\cdot\vec{e}_2#. What follows, after subtracting #v_2\vec{e}_2-w_1\cdot\vec{e}_1# on both sides:\[\left(v_1-w_1\right)\cdot \vec{e}_1 = \left(w_2-v_2\right)\cdot \vec{e}_2\] Because #\vec{e}_1# and #\vec{e}_2# are not on the same line as #\vec{0}#, we must conclude with the *Criteria for two vectors on a line through the origin* that #v_1-w_1=0# and #w_2-v_2=0#, in other words #v_1=w_1# and #v_2=w_2#. This shows that the coordinates #v_1# and #v_2# are unique.

The same goes for the space: say \[v_1\cdot \vec{e}_1+v_2\cdot\vec{e}_2+v_3\cdot\vec{e}_3=w_1\cdot\vec{e}_1+w_2\cdot\vec{e}_2+w_3\cdot\vec{e}_3\] This can be rewritten as \[\left(v_1-w_1\right)\cdot \vec{e}_1 + \left(v_2-w_2\right)\cdot \vec{e}_2+ \left(v_3-w_3\right)\cdot\vec{e}_3=0\] If #v_1-w_1\ne0#, then after multiplying both sides with the the scalar #\frac{1}{v_1-w_1}#, we see that #\vec{e}_1# is a linear combination of #\vec{e}_2# and #\vec{e}_3#; this means that #\vec{e}_1# is in the plane through #\vec{0}#, #\vec{e}_2#, and #\vec{e}_3#, a contradiction to the assumption. What should apply is: #v_1=w_1#. With the same reasoning for the indices #2# and #3#, rather than #1#, we deduce that #v_2=w_2# and #v_3=w_3#. This shows that the coordinates #v_1#, #v_2#, and #v_3# are unique.

We next discuss how to recognize a basis.

Two characteristics of a basis

We fix an origin of the plane and the space.

- Two vectors in the plane, respectively three vectors in the space, form a basis if and only if each vector of the plane, respectively the space, is a linear combination of these.
- Two vectors in the plane, respectively three vectors in the space, form a basis if and only if the only way to write the origin as a linear combination of these vectors, is when all scalars are equal to #0#.

To determine this, we use the second

*characteristic of a basis*. Suppose there are two scalars #\lambda# and #\mu#, so #\vec{0}=\lambda\cdot \left[ -2 , 5 \right]+\mu\cdot \left[ -6 , 15 \right]#. #\lambda# and #\mu# satisfy the system of linear equations \[\eqs{ -2\cdot \lambda-6\cdot \mu&=&0\cr 5\cdot \lambda+15\cdot \mu&=&0}\] whose solutions are: #\lambda= -3 \mu#. In particular, #\lambda= -6\land \mu=2# is a solution.

This shows that #\vec{0}# can be written as the linear combination \[ \lambda \cdot \left[ -2 , 5 \right]+\mu\cdot \left[ -6 , 15 \right]\] wherein #\lambda= -6\land \mu=2# are scalars that are not both equal to #0#. The second

*characteristic of a basis*gives that #\left[ -2 , 5 \right]# and #\left[ -6 , 15 \right]# do not form a basis.

The first

*characteristic of a basis*can also be used to derive this. If #\vec{v}=\rv{v_1,v_2}# is a linear combination #\vec{v}=\lambda\cdot{ \left[ -2 , 5 \right]}+\mu\cdot {\left[ -6 , 15 \right]}# of #\left[ -2 , 5 \right]# and #\left[ -6 , 15 \right]#, with scalars #\lambda # and #\mu#, those for the scalars satisfy the system of linear equations \[\eqs{ -2\cdot \lambda-6\cdot \mu&=&v_1\cr 5\cdot \lambda+15\cdot \mu&=&v_2 }\] This shows that the vector #\vec{v}=\rv{1,0} # cannot be written as a linear combination of #\left[ -2 , 5 \right]# and #\left[ -6 , 15 \right]#. The first

*characteristic of a basis*gives that #\left[ -2 , 5 \right]# and #\left[ -6 , 15 \right]# do not form a basis.

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