A vector is a quantity that has both a magnitude and a direction. Vectors can be visualized as arrows. The following figure shows what we mean by the components
of a vector
The sum
of two vectors has the components
- Explain the addition of vectors in terms of arrows.
The dot product of two vectors is the number
![{\displaystyle \mathbf {a} \cdot \mathbf {b} =a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e85642c71ae4bd35f4c02ecbf4887da7ea1bc995)
Its importance arises from the fact that it is invariant under rotations. To see this, we calculate
![{\displaystyle (\mathbf {a} +\mathbf {b} )\cdot (\mathbf {a} +\mathbf {b} )=(a_{x}+b_{x})^{2}+(a_{y}+b_{y})^{2}+(a_{z}+b_{z})^{2}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/339dc9ddf3b3fbe8b88f896f3ef62dc627c0c09b)
![{\displaystyle a_{x}^{2}+a_{y}^{2}+a_{z}^{2}+b_{x}^{2}+b_{y}^{2}+b_{z}^{2}+2\,(a_{x}b_{x}+a_{y}b_{y}+a_{z}b_{z})=\mathbf {a} \cdot \mathbf {a} +\mathbf {b} \cdot \mathbf {b} +2\,\mathbf {a} \cdot \mathbf {b} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63c73dbec128dc025da54e00096f0daa5451d7aa)
According to Pythagoras, the magnitude of
is
If we use a different coordinate system, the components of
will be different:
But if the new system of axes differs only by a rotation and/or translation of the axes, the magnitude of
will remain the same:
![{\displaystyle {\sqrt {a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}={\sqrt {(a'_{x})^{2}+(a'_{y})^{2}+(a'_{z})^{2}}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e41691d9b874f2ed677de664044b1e70d923d563)
The squared magnitudes
and
are invariant under rotations, and so, therefore, is the product
- Show that the dot product is also invariant under translations.
Since by a scalar we mean a number that is invariant under certain transformations (in this case rotations and/or translations of the coordinate axes), the dot product is also known as (a) scalar product. Let us prove that
![{\displaystyle \mathbf {a} \cdot \mathbf {b} =ab\cos \theta ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eff890657e46a885f9ddf58d5262e51ac171f80c)
where
is the angle between
and
To do so, we pick a coordinate system
in which
In this coordinate system
with
Since
is a scalar, and since scalars are invariant under rotations and translations, the result
(which makes no reference to any particular frame) holds in all frames that are rotated and/or translated relative to
We now introduce the unit vectors
whose directions are defined by the coordinate axes. They are said to form an orthonormal basis. Ortho because they are mutually orthogonal:
![{\displaystyle \mathbf {\hat {x}} \cdot \mathbf {\hat {y}} =\mathbf {\hat {x}} \cdot \mathbf {\hat {z}} =\mathbf {\hat {y}} \cdot \mathbf {\hat {z}} =0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f06424738f41173958de84d62fdb4f864db129d1)
Normal because they are unit vectors:
![{\displaystyle \mathbf {\hat {x}} \cdot \mathbf {\hat {x}} =\mathbf {\hat {y}} \cdot \mathbf {\hat {y}} =\mathbf {\hat {z}} \cdot \mathbf {\hat {z}} =1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6d7a85fbfb63c60d43ac1496fc7d6d9eb5aaf24)
And basis because every vector
can be written as a linear combination of these three vectors — that is, a sum in which each basis vector appears once, multiplied by the corresponding component of
(which may be 0):
![{\displaystyle \mathbf {v} =v_{x}\mathbf {\hat {x}} +v_{y}\mathbf {\hat {y}} +v_{z}\mathbf {\hat {z}} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9509a8be1a1e0c4cb6a35b04e307a237259f4e67)
It is readily seen that
which is why we have that
![{\displaystyle \mathbf {v} =\mathbf {\hat {x}} \,(\mathbf {\hat {x}} \cdot \mathbf {v} )+\mathbf {\hat {y}} \,(\mathbf {\hat {y}} \cdot \mathbf {v} )+\mathbf {\hat {z}} \,(\mathbf {\hat {z}} \cdot \mathbf {v} ).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d42776cd0b4982c394406ce1d7327dade4ffc23a)
Another definition that is useful (albeit only in a 3-dimensional space) is the cross product of two vectors:
![{\displaystyle \mathbf {a} \times \mathbf {b} =(a_{y}b_{z}-a_{z}b_{y})\,\mathbf {\hat {x}} +(a_{z}b_{x}-a_{x}b_{z})\,\mathbf {\hat {y}} +(a_{x}b_{y}-a_{y}b_{x})\,\mathbf {\hat {z}} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2800d90a015d06567e82d6862bf1ae6820eeed89)
- Show that the cross product is antisymmetric:
![{\displaystyle \mathbf {a} \times \mathbf {b} =-\mathbf {b} \times \mathbf {a} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/56476ccda6abb8cabfb63441fc43f810fa987df3)
As a consequence,
- Show that
![{\displaystyle \mathbf {a} \cdot (\mathbf {a} \times \mathbf {b} )=\mathbf {b} \cdot (\mathbf {a} \times \mathbf {b} )=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27234a9d3490c77694f1866a68120234e117a9e4)
Thus
is perpendicular to both
and
- Show that the magnitude of
equals
where
is the angle between
and
Hint: use a coordinate system in which
and ![{\displaystyle \mathbf {b} =(b\cos \alpha ,b\sin \alpha ,0).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0ea23b799e2c89328f883f8877178bdaf0278b76)
Since
is also the area
of the parallelogram
spanned by
and
we can think of
as a vector of magnitude
perpendicular to
Since the cross product yields a vector, it is also known as vector product.
(We save ourselves the trouble of showing that the cross product is invariant under translations and rotations of the coordinate axes, as is required of a vector. Let us however note in passing that if
and
are polar vectors, then
is an axial vector. Under a reflection (for instance, the inversion of a coordinate axis) an ordinary (or polar) vector is invariant, whereas an axial vector changes its sign.)
Here is a useful relation involving both scalar and vector products:
![{\displaystyle \mathbf {a} \times (\mathbf {b} \times \mathbf {c} )=\mathbf {b} (\mathbf {c} \cdot \mathbf {a} )-(\mathbf {a} \cdot \mathbf {b} )\mathbf {c} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ee19134339415e9d677e0fbf0b4e6c8460d798f)