1107: For Real Vectors Space with Inner Product and 2 Linearly Independent Unit Vectors with Angle θ, Unit Vector on Plane with Angle θ′ from 1st Vector Is This

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description/proof of that for real vectors space with inner product and $2$ linearly independent unit vectors with angle $\theta$, unit vector on plane with angle ${\theta }^{\prime }$ from 1st vector is this

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note 1
• 3: Proof
• 4: Note 2

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Starting Context

• The reader knows a definition of inner product on real or complex vectors space.

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Target Context

• The reader will have a description and a proof of the proposition that for any real vectors space with any inner product and any $2$ linearly independent unit vectors with any angle $\theta$, the unit vector on the plane with any angle ${\theta }^{\prime }$ from the 1st vector is this.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$\mathbb{R}$: with the canonical field structure
$V$: $\in \{\text{ the }\mathbb{R}\text{ vectors spaces }\}$ with any inner product, $⟨\bullet ,\bullet ⟩$
$u_1$: $\in V$, with $⟨u_1,u_1⟩=1$
$u_2$: $\in V$, with $⟨u_2,u_2⟩=1$
$u$: $\in V$, $=c{u}_1+d{u}_2$, with $⟨u,u⟩=1$
$\theta$: $\in \mathbb{R}$ such that $0<\theta <\pi$
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Statements:
(
$⟨u_1,u_2⟩=cos\theta$
$\wedge$
$⟨u_1,u⟩=cos{\theta }^{\prime }$ where $0\le {\theta }^{\prime }<2\pi$
$\wedge$
$⟨u_2,u⟩=cos({\theta }^{\prime }-\theta )$
)
$⟹$
$c=cos{\theta }^{\prime }-(sin{\theta }^{\prime }/sin\theta )cos\theta \wedge d=sin{\theta }^{\prime }/sin\theta$
//

$⟨u_2,u⟩=cos({\theta }^{\prime }-\theta )$ instead of $cos({\theta }^{\prime }+\theta )$ means nothing but that ${\theta }^{\prime }$ is taken in the same direction with $\theta$.

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2: Note 1

The dimension of $V$ can be any including infinity.

In other words, $u$ is $u_1$ rotated ${\theta }^{\prime }$ toward $u_2$.

On any $d$-dimensional real vectors space, a rotation around a vector as "axis" is a '$d-1$'-dimensional rotation; on the '$d-1$'-dimensional subspace, a rotation around a vector as "axis" is a '$d-2$'-dimensional rotation; ...; on the $3$-dimensional subspace, a rotation around a vector as "axis" is a $2$-dimensional rotation.

So, any $2$-dimensional rotation on any $d$-dimensional real vectors space requires some $d-2$ axes.

When one thinks of a $2$-dimensional rotation on a $d$-dimensional real vectors space, often, it is simpler to just think of the $2$-dimensional subspace than to think of the $d-2$ axes.

This proposition is thinking of the $2$-dimensional subspace spanned by $u_1$ and $u_2$.

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3: Proof

Whole Strategy: Step 1: put $u=c{u}_1+d{u}_2$ into $⟨u_1,u⟩=cos{\theta }^{\prime }$ and $⟨u,u⟩=1$, and get the candidates for $c$ and $d$; Step 2: choose the unique $c$ and $d$.

Step 1:

Let us put $u=c{u}_1+d{u}_2$ into $⟨u_1,u⟩=cos{\theta }^{\prime }$ and $⟨u,u⟩=1$.

$cos{\theta }^{\prime }=⟨u_1,c{u}_1+d{u}_2⟩=c⟨u_1,u_1⟩+d⟨u_1,u_2⟩=c+dcos\theta$.

$1=⟨u,u⟩=⟨c{u}_1+d{u}_2,c{u}_1+d{u}_2⟩=c^2⟨u_1,u_1⟩+cd⟨u_1,u_2⟩+dc⟨u_2,u_1⟩+d^2⟨u_2,u_2⟩=c^2+2cdcos\theta +d^2$.

As $c=cos{\theta }^{\prime }-dcos\theta$, $1=(cos{\theta }^{\prime }-dcos\theta {)}^2+2(cos{\theta }^{\prime }-dcos\theta )dcos\theta +d^2=d^2(1+co{s}^2\theta -2co{s}^2\theta )+d(-2cos{\theta }^{\prime }cos\theta +2cos{\theta }^{\prime }cos\theta )+co{s}^2{\theta }^{\prime }=d^2(1-co{s}^2\theta )+co{s}^2{\theta }^{\prime }=d^{2}si{n}^2\theta +co{s}^2{\theta }^{\prime }$, so, $d^2=(1-co{s}^2{\theta }^{\prime })/si{n}^2\theta =si{n}^2{\theta }^{\prime }/si{n}^2\theta$ ($sin\theta \ne 0$ because $0<\theta <\pi$), so, $d=+-sin{\theta }^{\prime }/sin\theta$.

$c=cos{\theta }^{\prime }-+(sin{\theta }^{\prime }/sin\theta )cos\theta$.

Step 2:

While Step 1 has gotten only candidates, $c$ and $d$ should be uniquely determined, because as ${\theta }^{\prime }$ is dictated to be unambiguously taken in the same direction with $\theta$, $u$ should be uniquely determined.

In other words, the condition, $⟨u_2,u⟩=cos({\theta }^{\prime }-\theta )$, should determine $c$ and $d$ uniquely.

$⟨u_2,u⟩=⟨u_2,c{u}_1+d{u}_2⟩=c⟨u_2,u_1⟩+d⟨u_2,u_2⟩=ccos\theta +d=(cos{\theta }^{\prime }-+(sin{\theta }^{\prime }/sin\theta )cos\theta )cos\theta +-sin{\theta }^{\prime }/sin\theta =cos{\theta }^{\prime }cos\theta -+(sin{\theta }^{\prime }/sin\theta )co{s}^2\theta +-sin{\theta }^{\prime }/sin\theta =cos{\theta }^{\prime }cos\theta +-(sin{\theta }^{\prime }/sin\theta -(sin{\theta }^{\prime }/sin\theta )co{s}^2\theta )=cos{\theta }^{\prime }cos\theta +-sin{\theta }^{\prime }/sin\theta (1-co{s}^2\theta )=cos{\theta }^{\prime }cos\theta +-sin{\theta }^{\prime }/sin\theta (si{n}^2\theta )=cos{\theta }^{\prime }cos\theta +-sin{\theta }^{\prime }sin\theta$.

On the other hand, $cos({\theta }^{\prime }-\theta )=cos(-\theta )cos{\theta }^{\prime }-sin(-\theta )sin{\theta }^{\prime }=cos\theta cos{\theta }^{\prime }+sin\theta sin{\theta }^{\prime }$.

So, $c=cos{\theta }^{\prime }-(sin{\theta }^{\prime }/sin\theta )cos\theta$ and $d=sin{\theta }^{\prime }/sin\theta$ is the answer.

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4: Note 2

Especially, when $\theta =\pi /2$, which means that $(u_1,u_2)$ is orthonormal, $c=cos{\theta }^{\prime }$ and $d=sin{\theta }^{\prime }$, and $u=cos{\theta }^{\prime }{u}_1+sin{\theta }^{\prime }{u}_2$.

For any vector on the $2$-dimensional subspace, it can be expressed as $u=r{u}_1=r({u_1}^1{e}_1+{u_1}^2{e}_2)$ for any orthonormal basis for the subspace, $(e_1,e_2)$. $u_2$ can be taken to be $u_2=-{u_1}^2{e}_1+{u_1}^1{e}_2$ to be orthonormal to $u_1$. By this proposition, $u_1$ is rotated ${\theta }^{\prime }$ to be $cos{\theta }^{\prime }{u}_1+sin{\theta }^{\prime }{u}_2=cos{\theta }^{\prime }({u_1}^1{e}_1+{u_1}^2{e}_2)+sin{\theta }^{\prime }(-{u_1}^2{e}_1+{u_1}^1{e}_2)=(cos{\theta }^{\prime }{u_1}^1-sin{\theta }^{\prime }{u_1}^2)e_1+(cos{\theta }^{\prime }{u_1}^2+sin{\theta }^{\prime }{u_1}^1)e_2$, and $u$ is rotated ${\theta }^{\prime }$ to be $r((cos{\theta }^{\prime }{u_1}^1-sin{\theta }^{\prime }{u_1}^2)e_1+(cos{\theta }^{\prime }{u_1}^2+sin{\theta }^{\prime }{u_1}^1)e_2)=((cos{\theta }^{\prime }r{u_1}^1-sin{\theta }^{\prime }r{u_1}^2)e_1+(cos{\theta }^{\prime }r{u_1}^2+sin{\theta }^{\prime }r{u_1}^1)e_2)$, which means that the transformation matrix for the components with respect to the basis is $\left(\begin{array}{cc}cos{\theta }^{\prime }& -sin{\theta }^{\prime }\\ sin{\theta }^{\prime }& cos{\theta }^{\prime }\end{array}\right)$.

For any vector, $v\in V$, $v=u+v^{\prime }$, where $v^{\prime }$ is orthogonal to $e_1$ and $e_2$, and $v$ is ${\theta }^{\prime }$ rotated to be $((cos{\theta }^{\prime }r{u_1}^1-sin{\theta }^{\prime }r{u_1}^2)e_1+(cos{\theta }^{\prime }r{u_1}^2+sin{\theta }^{\prime }r{u_1}^1)e_2)+v^{\prime }$.

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References

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