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.
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#.
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}#.
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.
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.
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.
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[ 6 , 5 , -4 \right]# with respect to the basis #\left[ 6 , -6 , -4 \right]#, #\left[ 6 , -6 , -5 \right]#, and #\left[ 1 , 0 , 0 \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[ -{{49}\over{6}} , {{22}\over{3}} , 11 \right]#
The coordinates we are looking for, are the trio #\rv{v_1,v_2,v_3}#, so \[\left[ 6 , 5 , -4 \right]= v_1\cdot \left[ 6 , -6 , -4 \right] + v_2\cdot \left[ 6 , -6 , -5 \right]+v_3\cdot \left[ 1 , 0 , 0 \right]\] We will solve the following system of linear equations with unknown #v_1#, #v_2#, and #v_3#: \[\eqs{ v_{3}+6\cdot v_{2}+6\cdot v_{1}&=&6\cr -6\cdot v_{2}-6\cdot v_{1}&=&5\cr -5\cdot v_{2}-4\cdot v_{1}&=&-4\cr }\] The only solution is # v_{1}=-{{49}\over{6}} \land v_{2}={{22}\over{3}} \land v_{3}=11
#. This shows that the coordinates of #\left[ 6 , 5 , -4 \right]# with respect to the given basis are equal to \[\left[ -{{49}\over{6}} , {{22}\over{3}} , 11 \right]\]
Coordinate space
