Given the 3D vectors
, whose coordinates are given for a Cartesian coordinate system:
,
,
their dot product may be defined to be
![{\displaystyle {\vec {u}}\cdot {\vec {v}}=u_{x}v_{x}+u_{y}v_{y}+u_{z}v_{z}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbda3fb53fa5c12fe698dd452403dd1f731be23d)
Let
be their cross product:
.
The cross product is perpendicular to both of factors,
and
, as can be verified by taking dot products:
![{\displaystyle =u_{x}u_{y}v_{z}-u_{x}u_{z}v_{y}+u_{y}u_{z}v_{x}-u_{y}u_{x}v_{z}+u_{z}u_{x}v_{y}-u_{z}u_{y}v_{x}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b08230f4fa0906652c880ca73353785549f47a1f)
Assuming that both
and
are non-zero in length then the cosine of the angle between them yields 0; so the angle between them is a right angle; so
and
are perpendicular; i.e.,
and
are perpendicular. [But the geometric interpretation for the 3D case has not been shown yet; would not this claim be circular then?]
Exercise: Show that
and
are perpendicular.
Let
be the normalized version of
, i.e.
![{\displaystyle {\hat {w}}={{\vec {w}} \over {\sqrt {{\vec {w}}\cdot {\vec {w}}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45351d4b42bf7ca56ffeeb1d41318e0ec8da627f)
so that
has unit length, because
![{\displaystyle {\hat {w}}\cdot {\hat {w}}={{\vec {w}}\cdot {\vec {w}} \over {\sqrt {{\vec {w}}\cdot {\vec {w}}}}^{2}}={{\vec {w}}\cdot {\vec {w}} \over {\vec {w}}\cdot {\vec {w}}}=1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05356c3c4fdd8fc3f2ae32228a9f5119b1d2e0fd)
and
. [Again, this calculation has used the geometric interpretation of the dot product already.]
Let
,
then
has the same length as
but is perpendicular to both
and
, so it is a rotation by 90° of
in the “
plane” (the plane whose normal is
).
Likewise let
.
Let
,
.
Note:
is a rotation of
by an angle
around axis
. Likewise
is a rotation of
by an angle
around axis
.
![{\displaystyle =\cos ^{2}\theta \,{\vec {u}}\cdot {\vec {v}}+\cos \theta \sin \theta \,{\vec {u}}\cdot {\vec {v}}_{\perp }+\sin \theta \cos \theta \,{\vec {u}}_{\perp }\cdot {\vec {v}}+\sin ^{2}\theta \,{\vec {u}}_{\perp }\cdot {\vec {v}}_{\perp }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ba1d49eb2413f6ead6dfefb776478a00af5da9b)
![{\displaystyle =\cos ^{2}\theta \,{\vec {u}}\cdot {\vec {v}}+\sin ^{2}\theta \,({\hat {w}}\times {\vec {u}})\cdot ({\hat {w}}\times {\vec {v}})+\cos \theta \sin \theta \,({\vec {u}}\cdot {\hat {w}}\times {\vec {v}}+{\hat {w}}\times {\vec {u}}\cdot {\vec {v}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/101ae806e2f029a9a871116c807264d8bbcbc31c)
![{\displaystyle {\hat {u}}\cdot {\hat {w}}\times {\vec {v}}={\begin{vmatrix}u_{x}&u_{y}&u_{z}\\w_{x}&w_{y}&w_{z}\\v_{x}&v_{y}&v_{z}\end{vmatrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bececcd58caab05713480c34cef086b4055abfce)
(
get replaced by
due to the dot product; this combination of dot and cross product forming a determinant is called a vector triple product).
![{\displaystyle {\hat {w}}\times {\vec {u}}\cdot {\vec {v}}={\vec {v}}\cdot {\hat {w}}\times {\vec {u}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5c35d93a910b1a881d2a1f688ada5eb1d23a08ce)
because the dot product is commutative.
![{\displaystyle =-u\cdot {\hat {w}}\times {\vec {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6ffacf036316cfd13689a3dce9ffc88fe29d8fa)
because there is an interchange of rows one and three in the determinant; so swapping two factors of a vector triple product changes its sign.
So
,
![{\displaystyle {\vec {u}}'\cdot {\vec {v}}'=\cos ^{2}\theta \,{\vec {u}}\cdot {\vec {v}}+\sin ^{2}\theta \,{\vec {u}}_{\perp }\cdot {\vec {v}}_{\perp }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/256e3c176644fe3ec4e8b099c8658f523ea3ae7d)
Focusing on the last term (of the right-hand side (RHS)):
![{\displaystyle ={1 \over w^{2}}({\vec {w}}\times {\vec {u}})\cdot ({\vec {w}}\times {\vec {v}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/028c8dd0040416826a9394c92965358325e818d4)
![{\displaystyle ={1 \over w^{2}}(({\vec {u}}\times {\vec {v}})\times {\vec {u}})\cdot (({\vec {u}}\times {\vec {v}})\times {\vec {v}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/282b26de802f9c73877f9d584c43aea73e55e436)
Summary of the argument coming up ahead:
is a 90° (counterclockwise) rotation of
in the
plane and
is a 90° (counterclockwise) rotation of
in the
plane, so
![{\displaystyle {\vec {u}}_{\perp }\cdot {\vec {v}}_{\perp }={\vec {u}}\cdot {\vec {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4a15b72114418a5d4dfb81ef3280026946b909c)
thus
![{\displaystyle =(\cos ^{2}\theta +\sin ^{2}\theta )\,{\vec {u}}\cdot {\vec {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a1ae0a1834747b54add4b73a571406b9a24fedc4)
(End of summary.)
The products
and
have the general form
. We will now derive a formula for
.
The double cross product
has to be perpendicular to
, so it must be in the plane of both
and
; it must be a linear combination of
and
.
![{\displaystyle =(b_{y}c_{z}-b_{z}c_{y},\,b_{z}c_{x}-b_{x}c_{z},\,b_{x}c_{y}-b_{y}c_{x})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a7ccf162c105426d7f6e68c3ec7ede0ba1e73b78)
![{\displaystyle a_{z}b_{y}c_{z}-a_{z}b_{z}c_{y}-a_{x}b_{x}c_{y}+a_{x}b_{y}c_{x},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/44477b067ea6c021d1801f0279ee3d472034be5b)
![{\displaystyle a_{x}b_{z}c_{x}-a_{x}b_{x}c_{z}-a_{y}b_{y}c_{z}+a_{y}b_{z}c_{y})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/395ec426b342cf8a47e3546fc3f51def38bdf40e)
![{\displaystyle b_{y}a_{x}c_{x}+b_{y}a_{z}c_{z}-c_{y}a_{x}b_{x}-c_{y}a_{z}b_{z},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba28bcb416aa81bbf1b25fe07ba31f0f5e24771a)
![{\displaystyle b_{z}a_{x}c_{x}+b_{z}a_{y}c_{y}-c_{z}a_{x}b_{x}-c_{z}a_{y}b_{y})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e51ef1814ddc12426aa01c2672c4f510b238696)
![{\displaystyle b_{y}(a_{x}c_{x}+a_{z}c_{z})-c_{y}(a_{x}b_{x}+a_{z}b_{z}),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8c3bf75927b1f66b460b1dc4b83e13d038a4bfe)
![{\displaystyle b_{z}(a_{x}c_{x}+a_{y}c_{y})-c_{z}(a_{x}b_{x}+a_{y}b_{y}))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a6d44e45eecae275e17f775025b8e7556bb1fd05)
![{\displaystyle b_{y}(a_{x}c_{x}+{\underline {a_{y}c_{y}}}+a_{z}c_{z})-c_{y}(a_{x}b_{x}+{\underline {a_{y}b_{y}}}+a_{z}b_{z}),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/881bab05975a59580058a71fc121b1f4f5b44339)
![{\displaystyle b_{z}(a_{x}c_{x}+a_{y}c_{y}+{\underline {a_{z}c_{z}}})-c_{z}(a_{x}b_{x}+a_{y}b_{y}+{\underline {a_{z}c_{z}}}))}](https://wikimedia.org/api/rest_v1/media/math/render/svg/74e97d411e9568c114cc4612d14226e48daf8bf2)
![{\displaystyle b_{y}\,{\vec {a}}\cdot {\vec {c}}-c_{y}\,{\vec {a}}\cdot {\vec {b}},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a0f954d412b464c1712f64694cfdf3f4977470ee)
![{\displaystyle b_{z}\,{\vec {a}}\cdot {\vec {c}}-c_{z}\,{\vec {a}}\cdot {\vec {b}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02893eb4e92385613bf6a57a97d2a6bd9fd15125)
![{\displaystyle =(b_{x}\,{\vec {a}}\cdot {\vec {c}},\,b_{y}\,{\vec {a}}\cdot {\vec {c}},\,b_{z}\,{\vec {a}}\cdot {\vec {c}})-(c_{x}\,{\vec {a}}\cdot {\vec {b}},\,c_{y}\,{\vec {a}}\cdot {\vec {b}},\,c_{z}\,{\vec {a}}\cdot {\vec {b}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12e7b5f0fdc580933ecbdd7b0c6fa34ca0df9016)
![{\displaystyle ={\vec {a}}\cdot {\vec {c}}\,(b_{x},\,b_{y},\,b_{z})-{\vec {a}}\cdot {\vec {b}}\,(c_{x},\,c_{y},\,c_{z})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/72047ecc36a30dcbab99dc74ab8c340df71c6c8e)
.
Back to the focused-on equation:
![{\displaystyle =-[({\vec {u}}\cdot {\vec {v}}){\vec {u}}-({\vec {u}}\cdot {\vec {u}}){\vec {v}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0741446f7a1bf502cf1a3607f375a22fc70b3fbb)
![{\displaystyle =u^{2}{\vec {v}}-({\vec {u}}\cdot {\vec {v}}){\vec {u}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a38c3cd1fcbd23a7791da1ca51cf24aba8bb00ad)
![{\displaystyle =-[({\vec {v}}\cdot {\vec {v}}){\vec {u}}-({\vec {v}}\cdot {\vec {u}}){\vec {v}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ffbfa1ceaac75f8a9f82406a65f33da28075d68)
![{\displaystyle =({\vec {v}}\cdot {\vec {u}}){\vec {v}}-v^{2}{\vec {u}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89cd00a62d6e8654c2a8eb87a319e174b8fa5f52)
Thus
![{\displaystyle ={1 \over w^{2}}\left\{(u^{2}{\vec {v}}-({\vec {u}}\cdot {\vec {v}}){\vec {u}})\cdot (({\vec {u}}\cdot {\vec {v}}){\vec {v}}-v^{2}{\vec {u}})\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e0a610c7a81e232ec224218d230a68d678f9984)
![{\displaystyle ={1 \over w^{2}}\left\{u^{2}({\vec {u}}\cdot {\vec {v}})v^{2}-u^{2}v^{2}({\vec {u}}\cdot {\vec {v}})-({\vec {u}}\cdot {\vec {v}})^{2}({\vec {u}}\cdot {\vec {v}})+({\vec {u}}\cdot {\vec {v}})v^{2}u^{2}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a39b96e670a6ffb6474a572f1ff0b1f02a5af5d)
![{\displaystyle ={1 \over w^{2}}\left\{u^{2}v^{2}({\vec {u}}\cdot {\vec {v}})-({\vec {u}}\cdot {\vec {v}})^{3}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05cd3900c6f133c0dce6511cecdaa03737c9ad09)
![{\displaystyle ={{\vec {u}}\cdot {\vec {v}} \over w^{2}}\left\{u^{2}v^{2}-({\vec {u}}\cdot {\vec {v}})^{2}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b10ca31d87eba85909ef35ce4aa696ae2882e3a0)
![{\displaystyle w^{2}={\vec {w}}\cdot {\vec {w}}=({\vec {u}}\times {\vec {v}})\cdot ({\vec {u}}\times {\vec {v}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c042829680c8febcebb1482dbb4c16e0eb70d6cf)
Now the first cross and the dot in the RHS may be swapped, because the product is a vector triple product:
![{\displaystyle w^{2}={\vec {u}}\cdot {\vec {v}}\times ({\vec {u}}\times {\vec {v}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a6d85c2371b0c6003cf404ed4d18528c035f04d)
and now there is a double cross product at the end of the RHS, so the formula can be applied:
![{\displaystyle ={\vec {u}}\cdot [v^{2}{\vec {u}}-({\vec {v}}\cdot {\vec {u}}){\vec {v}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e6b3e3098bc625eff0c938ed5ca6a774763f377)
![{\displaystyle =[v^{2}u^{2}-({\vec {v}}\cdot {\vec {u}})^{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e942e8f515e9318f7091380f205e926a0aa1ef3d)
Thus
![{\displaystyle {{\vec {u}}\cdot {\vec {v}} \over w^{2}}(u^{2}v^{2}-({\vec {u}}\cdot {\vec {v}})^{2})={\vec {u}}\cdot {\vec {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bd247c78787607b3a23009ff99ab797fa7b9c8b)
- ∴
![{\displaystyle {\vec {u}}_{\perp }\cdot {\vec {v}}_{\perp }={\vec {u}}\cdot {\vec {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f4a15b72114418a5d4dfb81ef3280026946b909c)
- ∴
.
So rotating 3D vectors
and
by the same angle along their common plane leaves their dot product preserved.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Given a pair of 3D vectors
and
, what happens to the dot product
if
is rotated around the axis
so that the angle between
and
is preserved?
Firstly we will derive the Rodrigues formula in order to perform such a rotation. Vector
must be analyzed into parts that are parallel and perpendicular to
: call them
and
. Firstly consider the normalized version of
:
.
The projection of
onto the
axis is:
![{\displaystyle {\vec {v}}_{\|}=({\vec {v}}\cdot {\hat {u}}){\hat {u}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4c8f1b299f267d39acc0d0acde16508aaee0f13)
Now subtract
from
to obtain its perpendicular part:
![{\displaystyle {\vec {v}}_{\perp }={\vec {v}}-{\vec {v}}_{\|}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2353d7b200a9bf60ec685432fcf46e675d63fef8)
What is
?
![{\displaystyle =({\vec {v}}-{\hat {u}}({\vec {v}}\cdot {\hat {u}}))\cdot {\hat {u}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e214e6ed06e7787c407c8ce9c3b23d8aff8ebee8)
![{\displaystyle ={\vec {v}}\cdot {\hat {u}}-({\hat {u}}\cdot {\hat {u}})({\vec {v}}\cdot {\hat {u}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af33cf3d138347dc760b03d0ff874aa0b93d3509)
Since
is a unit vector then
, so
![{\displaystyle {\vec {v}}_{\perp }\cdot {\hat {u}}={\vec {v}}\cdot {\hat {u}}-{\vec {v}}\cdot {\hat {u}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/113fb6eb0de73658127e187e69242f0503beab4f)
which proves that
is perpendicular to
, as expected. [Or does it? Has the geometric interpretation of the dot product been established yet? Is not the argument a bit circular?]
![{\displaystyle {\vec {v}}_{\perp }'={\hat {u}}\times {\vec {v}}_{\perp }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c7978b22030133733553f7a9b33605952e01c86)
![{\displaystyle {\vec {v}}'={\vec {v}}_{\|}+\cos \theta \,{\vec {v}}_{\perp }+\sin \theta \,{\vec {v}}'_{\perp }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2c94cc96ebc08f6a0bde4f610b032ae4d4b3e81)
where
is the angle of rotation.
![{\displaystyle ({\vec {v}}\cdot {\hat {u}}){\hat {u}}+\cos \theta ({\vec {v}}-({\vec {v}}\cdot {\hat {u}}){\hat {u}})+\sin \theta \,{\hat {u}}\times ({\vec {v}}-({\vec {v}}\cdot {\hat {u}}){\hat {u}})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cffe0688616db99f3f02913df50f692622de8cc)
![{\displaystyle ({\vec {v}}\cdot {\hat {u}}){\hat {u}}+\cos \theta \,({\vec {v}}-({\vec {v}}\cdot {\hat {u}}){\hat {u}})+\sin \theta \,{\hat {u}}\times {\vec {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a47f2a875e3f6ee142c452e8a71d90ad5a6230d)
so a Rodrigues rotation of vector
around vector
by an angle
is
![{\displaystyle R_{{\vec {u}},\,\theta }({\vec {v}}')=(1-\cos \theta )({\vec {v}}\cdot {\hat {u}}){\hat {u}}+\cos \theta \,{\vec {v}}+\sin \theta \,{\hat {u}}\times {\vec {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3e5220e15bc9ac8ed00c622fc40dc8976118274)
Dot-multiplying the just-above by
yields
![{\displaystyle {\vec {u}}\cdot R_{{\vec {u}},\,\theta }({\vec {v}}')=(1-\cos \theta )({\vec {v}}\cdot {\hat {u}})({\vec {u}}\cdot {\hat {u}})+\cos \theta \,{\vec {v}}\cdot {\vec {u}}+\sin \theta \,{\vec {u}}\cdot {\hat {u}}\times {\vec {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/414481ad55e899de020aa3bbfbd55e5333bd2e7d)
![{\displaystyle {\vec {u}}\cdot {\hat {u}}={\vec {u}}\cdot {{\vec {u}} \over u}={u^{2} \over u}=u}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ece8de2e2d9e0bc2102465ed295f6729cc05d10f)
![{\displaystyle {\vec {u}}\cdot {\hat {u}}\times {\vec {v}}={\vec {u}}\times {\hat {u}}\cdot {\vec {v}}={1 \over u}{\vec {u}}\times {\vec {u}}\cdot {\vec {v}}=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4ff2f8ff94cacd52339e948f5d8381aa69278236)
because
for any
.
![{\displaystyle =(1-\cos \theta )({\vec {v}}\cdot {\hat {u}}u)+\cos \theta \,{\vec {v}}\cdot {\vec {u}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed0ba3a852c861f5a052326f1355a2cb1b906135)
![{\displaystyle =(1-\cos \theta )({\vec {v}}\cdot {\vec {u}})+\cos \theta \,{\vec {v}}\cdot {\vec {u}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/14b5560469d5d6ef32f9868d1e3599438d3e2e0d)
![{\displaystyle ={\vec {v}}\cdot {\vec {u}}={\vec {u}}\cdot {\vec {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/008938c2ec2f36828387cbe51e24eb2abbc2deb5)
as claimed.
Any other rotations of a pair of vectors which leave the angle between them unchanged may be composed of these two kinds of rotation already considered: (1) Rodrigues rotation of one vector around the axis of the other vector, which also rotates the plane containing both vectors; and (2) a dual, “isoangular”, simultaneous pair of rotations of both vectors along their common plane, which leaves it fixed.
Both of these kinds of rotations have been shown to preserve the dot product between the two vectors; therefore any angle preserving (and magnitude preserving; but that should be implicit in the term “rotation”) rotational movement of the two vectors also preserves their dot product.
Homogeneity. Multiplying the lengths of
and
. Let
,
, i.e.,
![{\displaystyle {\vec {u}}'=(\lambda u_{x},\,\lambda u_{y},\,\lambda u_{z})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c45251d56c767decde95ddee35002e4d059eac4d)
,
then
.
Multiplying the length of
or
also multiplies the length of their dot product by the same factor.
What happens when both
and
are unit vectors? Rotate them in an angle-preserving way so as to place
along the x-axis. Then
.
Rotate
around
until
is contained by the xy-plane. Then
![{\displaystyle {\vec {v}}'=(v_{x}',\,v_{y}',\,0)=(\cos \phi ,\,\sin \phi ,\,0)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aaa92f6a421d84e119cce01c60fa1c01cd0edf8d)
for some angle
. The angle
is now contained within the xy-plane:
.
![{\displaystyle \cos \phi ={\vec {u}}\cdot {\vec {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4bbc96af703f39fce126c705c3f857d712b19db2)
because the rotation was also dot-product preserving.
For non-unit vectors
:
,
![{\displaystyle {\vec {v}}=v{\hat {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3857abe2a274b37bc12106fa3d7aa0e1a67f35f5)
where
,
.
![{\displaystyle {\vec {u}}\cdot {\vec {v}}=u{\hat {u}}\cdot v{\hat {v}}=uv\,{\hat {u}}\cdot {\hat {v}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d794097ced7929658631c583b02d210dfd7d9da2)
but
so
![{\displaystyle {\vec {u}}\cdot {\vec {v}}=uv\,\cos \phi }](https://wikimedia.org/api/rest_v1/media/math/render/svg/db2da1cd8bccd8b58a2e0fee33db30361ba311bb)
where u and v are the magnitudes of
respectively; and
is the angle between them. This is the geometric interpretation of the dot product (in 3D; it looks the same as that for the plane).