Vector calculus in plane and space: Distances, Angles and Inner Product
The standard dot product
Orthonormal basis
A basis of the plane or the space wherein each vector has a length #1#, and each pair is perpendicular to each other, is called an orthonormal basis.
The standard basiss of #\mathbb{R}^2# and #\mathbb{R}^3# are orthonormal.
We translate the dot product of #\vec{v}# and #\vec{w}# into coordinates relative to an orthonormal basis, to give an expression that is easy to remember.
The dot product in coordinates
If #\vec{e}_1#, #\vec{e}_2# is an orthonormal basis of the plane, the following applies to all vectors #\vec{v}# and #\vec{w}#:
\[
\vec{v}\cdot \vec{w} = v_1\cdot w_1 + v_2\cdot w_2\tiny,
\] wherein #\rv{v_1,v_2}# are the coordinates of #\vec{v}# regarding the basis #\vec{e}_1#, #\vec{e}_2#, and #\rv{w_1,w_2}# are the coordinates of #\vec{w}# regarding the same basis.
If #\vec{e}_1#, #\vec{e}_2#, #\vec{e}_3# is an orthonormal basis of the space, the following applies for all vectors #\vec{v}# and #\vec{w}#:
\[
\vec{v}\cdot \vec{w} = v_1\cdot w_1 + v_2\cdot w_2+ v_3\cdot w_3
\] where #\rv{v_1,v_2,v_3}# are the coordinates of #\vec{v}# regarding the basis #\vec{e}_1#, #\vec{e}_2#, #\vec{e}_3#, and #\rv{w_1,w_2,w_3}# are the coordinates of #\vec{w}# regarding the same basis.
We give a proof for the two-dimensional case. The three-dimensional case can be proved in the same way.
If #\vec{v}= v_1\vec{e}_1+v_2 \vec{e}_2# and #\vec{w}= w_1\vec{e}_1 +w_2 \vec{e}_2# are two vectors in the plane, we find, using the rules for calculating the dot product and the equalities #\vec{e}_1\cdot\vec{e}_1=1#, #\vec{e}_1\cdot\vec{e}_2=0#, #\vec{e}_2\cdot\vec{e}_1=0#, #\vec{e}_2\cdot\vec{e}_2=1#:
\[
\begin{array}{rcl}
\vec{v}\cdot \vec{w} & = &\left(v_1\vec{e}_1 +v_2 \vec{e}_2\right)\cdot\left(w_1\vec{e}_1 +w_2 \vec{e}_2\right) \\
& =& v_1\cdot w_1 \left(\vec{e}_1\cdot\vec{e}_1\right) + v_1\cdot w_2\cdot\left (\vec{e}_1\cdot \vec{e}_2\right)\\
&&+ v_2\cdot w_1\cdot\left (\vec{e}_2\cdot \vec{e}_1\right) + v_2\cdot w_2\cdot\left (\vec{e}_2\cdot\vec{e}_2\right)\\
& = &v_1\cdot w_1 + v_2\cdot w_2
\end{array}
\]
When the basis is orthonormal, it is more difficult to describe the coordinates; we have left this description out.
The standard basis of #\mathbb{R}^3# (as well as of #\mathbb{R}^2#) is orthonormal, which means that this formula applies in particular to the dot product in #\mathbb{R}^3# regarding said basis. To distinguish this dot product from dot products expressed in coordinates from other basis choices, we give it a special name:
Standard dot product
The standard dot product of two vectors #\vec{v}= \rv{v_1, v_2,v_3}# and #\vec{w}= \rv{w_1, w_2,w_3}# in #\mathbb{R}^3#, is the dot product expressed in terms of coordinates relative to the standard basis. It is given by the formula for 3 dimensions in the theory above, the dot product in coordinates.
The corresponding definition applies to #\mathbb{R}^2#.
We generally refer to the dot product when we mean standard dot product and there is no confusion.
As a result of its position, we find a convenient (and familiar) expression for the length of a vector. We formulate the result only for the space. The corresponding results for the plane can be read by setting the third coordinate of each vector equal to #0#.
Let #\vec{e}_1#, #\vec{e}_2#, #\vec{e}_3# be an orthonormal basis of the space, and look at two vectors #\vec{u}= u_1\vec{e}_1+u_2 \vec{e}_2+u_3 \vec{e}_3# and #\vec{v}= v_1\vec{e}_1+v_2 \vec{e}_2+v_3 \vec{e}_3#. We can express the length of the vector #\vec{v}#, the distance between #\vec{u} # and #\vec{v} #, and, if #\vec{u}# and #\vec{v}# are not the zero vector, the cosine of the angle #\varphi# between the two vectors, in coordinates as follows:
- #\parallel\vec{v}\parallel = \sqrt{v_1^2 + v_2^2+ v_3^2}#
- #\parallel\vec{u}-\vec{v}\parallel = \sqrt{\left(u_1 -v_1\right)^2 +\left (u_2 -v_2\right)^2+\left(u_3 -v_3\right)^2 }#
- #\cos (\varphi) =\frac{u_1\cdot v_1 + u_2\cdot v_2+u_3\cdot v_3}{\sqrt{u_1^2 + u_2^2+ u_3^2}\cdot \sqrt{v_1^2 + v_2^2+ v_3^2}}#
When using an orthonormal basis #\vec{e}_1#, #\vec{e}_2#, #\vec{e}_3# in the space (i.e., vectors of a length 1 which are perpendicular to each other two by two), we find the following expression in the coordinates for the dot product of the vectors #\vec{v}=v_1\vec{e}_1 +v_2 \vec{e}_2 + v_3 \vec{e}_3# and #\vec{w}=w_1\vec{e}_1 +w_2 \vec{e}_2 + w_3 \vec{e}_3#:
\[
v_1\cdot w_1 + v_2\cdot w_2 + v_3\cdot w_3\tiny.
\]
For the length of the vector #\vec{v}# we get
\[
\parallel\vec{v}\parallel =\sqrt{\vec{v}\cdot\vec{v}}=\sqrt{v_1^2 + v_2^2 + v_3^2}\tiny.
\]
The distance between #\vec{u}# and #\vec{v}# is equal to
\[
\parallel\vec{u}-\vec{v}\parallel = \sqrt{(u_1 -v_1)^2 + (u_2 -v_2)^2 + (u_3-v_3)^2}\tiny.
\]
And for the cosine of the angle between the vectors #\vec{v}# and #\vec{w}# (both are not equal #\vec{0}#, the following applies):
\[
\cos (\varphi) = \frac{\vec{v}\cdot \vec{w}}{\parallel\vec{v}\parallel\cdot \parallel\vec{w}\parallel} =
\frac{v_1\cdot w_1 + v_2\cdot w_2 + v_3\cdot w_3}{\sqrt{v_1^2 + v_2^2 + v_3^2}\cdot \sqrt{w_1^2 + w_2^2 + w_3^2}}\tiny.
\]
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