Coordinate space

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

Coordinate space

We have treated both vectors in the plane and in the space as vectors in the coordinate spaces $\mathbb{R}^2$ and $\mathbb{R}^3$ . Except for the choice of an origin and a basis, these concepts of vector in plane and space are the same as those in the coordinate spaces.

Coordinate space

Choose an origin $\vec{0}$ and a basis $\vec{e}_1$ , $\vec{e}_2$ , $\vec{e}_3$ of the space. Each vector $\vec{v}=v_1\cdot \vec{e}_1+v_2\cdot\vec{e}_2+v_3\cdot\vec{e}_3$ of $\mathbb{E}^3$ corresponds to a unique coordinate vector $\rv{v_1,v_2,v_3}$ in $\mathbb{R}^3$ . Written as a formula: $v_1\cdot \vec{e}_1+v_2\cdot\vec{e}_2+v_3\cdot\vec{e}_3\quad \leftrightarrow\quad \rv{v_1,v_2,v_3}$

The following rules are satisfied.

The origin $\vec{0}$ corresponds to $\rv{0,0,0}$ in $\mathbb{R}^3$ .
The basis $\vec{e}_1$ , $\vec{e}_2$ , $\vec{e}_3$ of the space corresponds to the standard basis $\rv{1,0,0}$ , $\rv{0,1,0}$ , $\rv{0,0,1}$ of $\mathbb{R}^3$ .
The addition of $\vec{w}=w_1\cdot \vec{e}_1+w_2\cdot\vec{e}_2+w_3\cdot\vec{e}_3$ in the space is linked to the addition in $\mathbb{R}^3$ by $\vec{v}+\vec{w} \leftrightarrow \rv{v_1+w_1 , v_2 + w_2 , v_3 + w_3}$ .
The scalar multiplication by $\lambda$ in the space is linked to the scalar multiplication in $\mathbb{R}^3$ by $\lambda \cdot \vec{v} \leftrightarrow \rv{\lambda \cdot v_1, \lambda \cdot v_2 , \lambda \cdot v_3}$ .

We call the set of all coordinate vectors the coordinate space, and write it as $\mathbb{R}^3$ .

Coordinates

The coordinate vector comprises the coordinates of $\vec{v}$ with respect to the given basis. The translation into coordinates depends on the choice of origin and basis!

We will deduce each of the four statements individually.

The origin $\vec{0}$ in the space is the linear combination $\vec{0} = 0\cdot \vec{e}_1+0\cdot\vec{e}_2+0\cdot\vec{e}_3$ , and, therefore, corresponds to $\rv{0,0,0}$ .
The basis vector $\vec{e}_1$ in the space is the linear combination $\vec{e}_1 = 1\cdot \vec{e}_1+0\cdot\vec{e}_2+0\cdot\vec{e}_3$ and, therefore, corresponds to $\rv{1,0,0}$ . A similar rule holds for $\vec{e}_2$ and $\vec{e}_3$ .
In order to add $\vec{v}$ and $\vec{w}$ , we use $\begin{array}{rcl}\vec{v}+\vec{w}&=&v_1\cdot \vec{e}_1+v_2\cdot\vec{e}_2+v_3\cdot\vec{e}_3+w_1\cdot \vec{e}_1+w_2\cdot\vec{e}_2+w_3\cdot\vec{e}_3\\&=&\left(v_1+w_1\right)\cdot \vec{e}_1+\left(v_2+w_2\right)\cdot\vec{e}_2+\left(v_3+w_3\right)\cdot\vec{e}_3\\ \end{array}$ The coordinate vector in $\mathbb{R}^3$ which corresponds to $\vec{v}+\vec{w}$ , is equal to $\rv{v_1+w_1,v_2+w_2,v_3+w_3}$ .
The scalar multiplication of $\lambda$ by $\vec{v}$ satisfies $\begin{array}{rcl}\lambda\cdot\vec{v}&=&\lambda\cdot \left(v_1\cdot\vec{e}_1+v_2\cdot\vec{e}_2+v_3\cdot\vec{e}_3\right)\\&=&\left(\lambda\cdot v_1\right)\cdot \vec{e}_1+\left(\lambda\cdot v_2\right)\cdot\vec{e}_2+\left(\lambda\cdot v_3\right)\cdot\vec{e}_3\\ \end{array}$ The coordinate vector in $\mathbb{R}^3$ which corresponds to $\lambda\cdot\vec{v}$ , is equal to $\rv{\lambda\cdot v_1,\lambda\cdot v_2,\lambda\cdot v_3}$ .

Plane

Similarly, $\mathbb{R}^2$ can play the role of coordinate plane in the Euclidean plane. The coordinate plane $\mathbb{R}^2$ and coordinate space $\mathbb{R}^3$ are primarily intended for the explicit calculation using numbers.

Correspondence

We can easily speak of the line in $\mathbb{R}^2$ or $\mathbb{R}^3$ , of vectors in $\mathbb{R}^2$ or $\mathbb{R}^3$ , the plane $\mathbb{R}^2$ , the space $\mathbb{R}^3$ , a parametric representation of a line in $\mathbb{R}^2$ , and so on. We could even write $\vec{a} = \rv{a_1 , a_2, a_3}$ if $\rv{a_1 , a_2, a_3}$ is the coordinate vector of $\vec{a}$ relative to a known origin and basis.

Higher dimensions

The use of $\mathbb{R}^n$ to describe the $n$ -dimensional space $\mathbb{E}^n$ goes through for $n\gt 3$ . We identify $\mathbb{R}^n$ with $\mathbb{E}^n$ after selection of an origin and basis in the latter. Although geometric objects such as points, lines, vectors, and planes, strictly speaking belong to the Euclidean space $\mathbb{E}^n$ , we can, therefore, view these objects as objects of the coordinate space $\mathbb{R}^n$ .

Later we will see that space and coordinate space are special cases of the notion of vector space.

Once we have identified the space $\mathbb{E}^3$ with $\mathbb{R}^3$ and we have specified a new origin and basis, then we can express the new coordinates in the old ones. We will give an example using a displacement of the origin.

Moving the origin in the coordinate space

Let $\vec{v}$ be a vector in the coordinate space $\mathbb{R}^3$ . If we move the origin and standard basis in $\mathbb{R}^3$ over the vector $\vec{v}$ (that is to say: add $\vec{v}$ ), the coordinate vector of a vector $\vec{w}$ , with respect to the moved basis, will be equal to $\vec{w}-\vec{v}$ .

The vector $\vec{w}$ represents the end point of the representative of the vector $\vec{w}$ whose starting point is in $\vec{0}$ . This point coincides with the end point of the representative of $\vec{w}-\vec{v}$ whose starting point is in $\vec{v}$ . If we choose the origin $\vec{v}$ , then $\vec{w}-\vec{v}$ is the vector whose representative with starting point indicates the original endpoint with $\vec{w}$ indicated point of the space.

After moving over $\vec{v}$ , the new origin will be in $\vec{v}$ and the new basis will be $\vec{e}_1+\vec{v}$ , $\vec{e}_2+\vec{v}$ , $\vec{e}_3+\vec{v}$ , wherein $\vec{e}_1$ , $\vec{e}_2$ , $\vec{e}_3$ is the standard basis of $\mathbb{R}^3$ . As we saw above, $\vec{e}_1=\left(\vec{v}+\vec{e}_1\right)-\vec{v}$ is the vector that, if it is set with starting point $\vec{v}$ has endpoint $\vec{v}+\vec{e}_1$ . With respect to the new origin, $\vec{e}_1$ corresponds to a vector of the standard basis; this also applies to $\vec{e}_2$ and $\vec{e}_3$ .

As we saw above, relative to this new basis, the coordinates of the point $\vec{w}=\rv{w_1,w_2,w_3}$ , are represented by the vector $\vec{w}-\vec{v}$ . This is also the coordinate vector of $\vec{w}$ relative to the standard basis: $\begin{array}{rcl}\vec{w}-\vec{v}&=&\rv{w_1-v_1,w_2-v_2,w_3-v_3}\\ &=&\left(w_1-v_1\right)\vec{e}_1+\left(w_2-v_2\right)\vec{e}_2+\left(w_3-v_3\right)\vec{e}_3\end{array}$

Determine the coordinates of the vector $\left[ 1 , -4 , -6 \right]$ with respect to the basis $\left[ 0 , 6 , 0 \right]$ , $\left[ -4 , 0 , 0 \right]$ , and $\left[ 0 , 0 , 1 \right]$ of $\mathbb{R}^3$ .

Give your answer in the form $\rv{v_1,v_2,v_3}$ , where $v_1$ , $v_2$ , and $v_3$ are integers.

$\left[ -{{2}\over{3}} , -{{1}\over{4}} , -6 \right]$

The coordinates we are looking for, are the trio $\rv{v_1,v_2,v_3}$ , so $\left[ 1 , -4 , -6 \right]= v_1\cdot \left[ 0 , 6 , 0 \right] + v_2\cdot \left[ -4 , 0 , 0 \right]+v_3\cdot \left[ 0 , 0 , 1 \right]$ We will solve the following system of linear equations with unknown $v_1$ , $v_2$ , and $v_3$ : $\eqs{ -4\cdot v_{2}&=&1\cr 6\cdot v_{1}&=&-4\cr v_{3}&=&-6\cr }$ The only solution is $v_{1}=-{{2}\over{3}} \land v_{2}=-{{1}\over{4}} \land v_{3}=-6$ . This shows that the coordinates of $\left[ 1 , -4 , -6 \right]$ with respect to the given basis are equal to $\left[ -{{2}\over{3}} , -{{1}\over{4}} , -6 \right]$

New example