How to show isometry of the space through plane?












0












$begingroup$


I am totally new to the isometries of the plane and space.
I have to prove that the map
$$R(vec{x})=vec{x}-2(vec{x}bullethat{n})hat{n}$$ from $R^{3}$ to $R^{3}$ is isometry of the space $R^{3}.$
The attempt what I have made is, I took two points $vec{x_{1}}$ and $vec{x_{2}}$ from $R^{3}$. The image of these two after the given map
$$R(vec{x_{1}})=vec{x_{1}}-2(vec{x_{1}}bullethat{n})hat{n}$$ and
$$R(vec{x_{2}})=vec{x_{2}}-2(vec{x_{2}}bullethat{n})hat{n}$$
Now I have to show that $R$ will be isometry if
$$||R(vec{x_{1}})-R(vec{x_{2}})||^{2}=||vec{x_{1}}-vec{x_{1}}||^{2}$$
SInce $R(vec{x_{1}})-R(vec{x_{2}})=(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}$
Now
$$||R(vec{x_{1}})-R(vec{x_{2}})||^{2}=big<(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}, (vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}big>$$
I get confused, how to proceed step by step to reach that
$$big<(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}, (vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}big>=||vec{x_{1}}-vec{x_{2}}||^{2}$$
$$Rightarrow~ bigg<vec{x_{1}}-vec{x_{2}},vec{x_{1}}-vec{x_{2}}bigg>-4big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg<vec{x_{1}}-vec{x_{2}},hat{n}bigg>+4bigg(big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg)^{2}bigg<hat{n},hat{n}bigg>$$
Am I right, how to show that $$4big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg<vec{x_{1}}-vec{x_{2}},hat{n}bigg>=4bigg(big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg)^{2}$$






isometry vector-space-isomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Use bilinearity of the scalar product.
    $endgroup$
    – amd
    Jan 16 at 17:56










  • $begingroup$
    @amd How can you explain sir!
    $endgroup$
    – Noor Aslam
    Jan 16 at 18:00










  • $begingroup$
    Recall the basic properties of inner products.
    $endgroup$
    – amd
    Jan 16 at 18:02










  • $begingroup$
    @amd can you explain in answer sir!
    $endgroup$
    – Noor Aslam
    Jan 16 at 18:03
















0












$begingroup$


I am totally new to the isometries of the plane and space.
I have to prove that the map
$$R(vec{x})=vec{x}-2(vec{x}bullethat{n})hat{n}$$ from $R^{3}$ to $R^{3}$ is isometry of the space $R^{3}.$
The attempt what I have made is, I took two points $vec{x_{1}}$ and $vec{x_{2}}$ from $R^{3}$. The image of these two after the given map
$$R(vec{x_{1}})=vec{x_{1}}-2(vec{x_{1}}bullethat{n})hat{n}$$ and
$$R(vec{x_{2}})=vec{x_{2}}-2(vec{x_{2}}bullethat{n})hat{n}$$
Now I have to show that $R$ will be isometry if
$$||R(vec{x_{1}})-R(vec{x_{2}})||^{2}=||vec{x_{1}}-vec{x_{1}}||^{2}$$
SInce $R(vec{x_{1}})-R(vec{x_{2}})=(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}$
Now
$$||R(vec{x_{1}})-R(vec{x_{2}})||^{2}=big<(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}, (vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}big>$$
I get confused, how to proceed step by step to reach that
$$big<(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}, (vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}big>=||vec{x_{1}}-vec{x_{2}}||^{2}$$
$$Rightarrow~ bigg<vec{x_{1}}-vec{x_{2}},vec{x_{1}}-vec{x_{2}}bigg>-4big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg<vec{x_{1}}-vec{x_{2}},hat{n}bigg>+4bigg(big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg)^{2}bigg<hat{n},hat{n}bigg>$$
Am I right, how to show that $$4big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg<vec{x_{1}}-vec{x_{2}},hat{n}bigg>=4bigg(big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg)^{2}$$






isometry vector-space-isomorphism







share|cite|improve this question











$endgroup$












  • $begingroup$
    Use bilinearity of the scalar product.
    $endgroup$
    – amd
    Jan 16 at 17:56










  • $begingroup$
    @amd How can you explain sir!
    $endgroup$
    – Noor Aslam
    Jan 16 at 18:00










  • $begingroup$
    Recall the basic properties of inner products.
    $endgroup$
    – amd
    Jan 16 at 18:02










  • $begingroup$
    @amd can you explain in answer sir!
    $endgroup$
    – Noor Aslam
    Jan 16 at 18:03














0












0








0





$begingroup$


I am totally new to the isometries of the plane and space.
I have to prove that the map
$$R(vec{x})=vec{x}-2(vec{x}bullethat{n})hat{n}$$ from $R^{3}$ to $R^{3}$ is isometry of the space $R^{3}.$
The attempt what I have made is, I took two points $vec{x_{1}}$ and $vec{x_{2}}$ from $R^{3}$. The image of these two after the given map
$$R(vec{x_{1}})=vec{x_{1}}-2(vec{x_{1}}bullethat{n})hat{n}$$ and
$$R(vec{x_{2}})=vec{x_{2}}-2(vec{x_{2}}bullethat{n})hat{n}$$
Now I have to show that $R$ will be isometry if
$$||R(vec{x_{1}})-R(vec{x_{2}})||^{2}=||vec{x_{1}}-vec{x_{1}}||^{2}$$
SInce $R(vec{x_{1}})-R(vec{x_{2}})=(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}$
Now
$$||R(vec{x_{1}})-R(vec{x_{2}})||^{2}=big<(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}, (vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}big>$$
I get confused, how to proceed step by step to reach that
$$big<(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}, (vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}big>=||vec{x_{1}}-vec{x_{2}}||^{2}$$
$$Rightarrow~ bigg<vec{x_{1}}-vec{x_{2}},vec{x_{1}}-vec{x_{2}}bigg>-4big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg<vec{x_{1}}-vec{x_{2}},hat{n}bigg>+4bigg(big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg)^{2}bigg<hat{n},hat{n}bigg>$$
Am I right, how to show that $$4big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg<vec{x_{1}}-vec{x_{2}},hat{n}bigg>=4bigg(big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg)^{2}$$






isometry vector-space-isomorphism







share|cite|improve this question











$endgroup$




I am totally new to the isometries of the plane and space.
I have to prove that the map
$$R(vec{x})=vec{x}-2(vec{x}bullethat{n})hat{n}$$ from $R^{3}$ to $R^{3}$ is isometry of the space $R^{3}.$
The attempt what I have made is, I took two points $vec{x_{1}}$ and $vec{x_{2}}$ from $R^{3}$. The image of these two after the given map
$$R(vec{x_{1}})=vec{x_{1}}-2(vec{x_{1}}bullethat{n})hat{n}$$ and
$$R(vec{x_{2}})=vec{x_{2}}-2(vec{x_{2}}bullethat{n})hat{n}$$
Now I have to show that $R$ will be isometry if
$$||R(vec{x_{1}})-R(vec{x_{2}})||^{2}=||vec{x_{1}}-vec{x_{1}}||^{2}$$
SInce $R(vec{x_{1}})-R(vec{x_{2}})=(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}$
Now
$$||R(vec{x_{1}})-R(vec{x_{2}})||^{2}=big<(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}, (vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}big>$$
I get confused, how to proceed step by step to reach that
$$big<(vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}, (vec{x_{1}}-vec{x_{2}})-2big((vec{x_{1}}-vec{x_{2}})bullethat{n}big)hat{n}big>=||vec{x_{1}}-vec{x_{2}}||^{2}$$
$$Rightarrow~ bigg<vec{x_{1}}-vec{x_{2}},vec{x_{1}}-vec{x_{2}}bigg>-4big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg<vec{x_{1}}-vec{x_{2}},hat{n}bigg>+4bigg(big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg)^{2}bigg<hat{n},hat{n}bigg>$$
Am I right, how to show that $$4big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg<vec{x_{1}}-vec{x_{2}},hat{n}bigg>=4bigg(big[(vec{x_{1}}-vec{x_{2}})bullethat{n}big]bigg)^{2}$$






isometry vector-space-isomorphism




isometry vector-space-isomorphism






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 13:28







Noor Aslam

















asked Jan 16 at 17:36









Noor AslamNoor Aslam

16012




16012












  • $begingroup$
    Use bilinearity of the scalar product.
    $endgroup$
    – amd
    Jan 16 at 17:56










  • $begingroup$
    @amd How can you explain sir!
    $endgroup$
    – Noor Aslam
    Jan 16 at 18:00










  • $begingroup$
    Recall the basic properties of inner products.
    $endgroup$
    – amd
    Jan 16 at 18:02










  • $begingroup$
    @amd can you explain in answer sir!
    $endgroup$
    – Noor Aslam
    Jan 16 at 18:03


















  • $begingroup$
    Use bilinearity of the scalar product.
    $endgroup$
    – amd
    Jan 16 at 17:56










  • $begingroup$
    @amd How can you explain sir!
    $endgroup$
    – Noor Aslam
    Jan 16 at 18:00










  • $begingroup$
    Recall the basic properties of inner products.
    $endgroup$
    – amd
    Jan 16 at 18:02










  • $begingroup$
    @amd can you explain in answer sir!
    $endgroup$
    – Noor Aslam
    Jan 16 at 18:03
















$begingroup$
Use bilinearity of the scalar product.
$endgroup$
– amd
Jan 16 at 17:56




$begingroup$
Use bilinearity of the scalar product.
$endgroup$
– amd
Jan 16 at 17:56












$begingroup$
@amd How can you explain sir!
$endgroup$
– Noor Aslam
Jan 16 at 18:00




$begingroup$
@amd How can you explain sir!
$endgroup$
– Noor Aslam
Jan 16 at 18:00












$begingroup$
Recall the basic properties of inner products.
$endgroup$
– amd
Jan 16 at 18:02




$begingroup$
Recall the basic properties of inner products.
$endgroup$
– amd
Jan 16 at 18:02












$begingroup$
@amd can you explain in answer sir!
$endgroup$
– Noor Aslam
Jan 16 at 18:03




$begingroup$
@amd can you explain in answer sir!
$endgroup$
– Noor Aslam
Jan 16 at 18:03










1 Answer
1






active

oldest

votes


















0












$begingroup$

Let us write $x$ instead of $vec{x}$ and $n$ instead of $hat{n}$. Moreover, $bullet$ denotes the usual dot product (scalar product) on Euclidean space $mathbb{R}^3$. In other words, $a bullet b = langle a, b rangle$ which is the alternative notation used in your question.



$R$ is a linear map because
$$R(x + y) = x + y - 2langle x + y, n rangle n = x + y - 2(langle x, n rangle + langle y, n rangle )n \= x - 2langle x, n rangle n + y - 2langle y, n rangle n = R(x) + R(y)$$
and
$$R(alpha x) = alpha x - 2langle alpha x, n rangle n = alpha x - 2alpha langle x, n rangle n = alpha R(x) .$$
It therefore suffices to show $lVert R(x) rVert =lVert x rVert$.
Simplifying your computations we get
$$lVert R(x) rVert^2 = langle R(x), R(x) rangle = langle x - 2langle x, n rangle n, x - 2langle x, n rangle n rangle \ = langle x, x rangle - 2langle x, n rangle langle n, x rangle - 2langle x, n rangle langle x, n rangle + 4langle x, n rangle^2 langle n, n rangle \ = lVert x rVert^2 - 4 langle x, n rangle ^2 + 4 langle x, n rangle ^2 lVert n rVert^2 .$$
You did not mention the assumption $lVert n rVert = 1$. But in that case you see that
$$lVert R(x) rVert^2 = lVert x rVert^2 .$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sir I have put hat on $n$ represents that $n$ is unit vector! SIr Thank You so much
    $endgroup$
    – Noor Aslam
    Jan 17 at 15:17










  • $begingroup$
    I see! But for future questions I recommend to explain precisely everything concerning notation. Otherwise you are in danger that some readers will not understand it correctly.
    $endgroup$
    – Paul Frost
    Jan 17 at 16:08












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

Let us write $x$ instead of $vec{x}$ and $n$ instead of $hat{n}$. Moreover, $bullet$ denotes the usual dot product (scalar product) on Euclidean space $mathbb{R}^3$. In other words, $a bullet b = langle a, b rangle$ which is the alternative notation used in your question.



$R$ is a linear map because
$$R(x + y) = x + y - 2langle x + y, n rangle n = x + y - 2(langle x, n rangle + langle y, n rangle )n \= x - 2langle x, n rangle n + y - 2langle y, n rangle n = R(x) + R(y)$$
and
$$R(alpha x) = alpha x - 2langle alpha x, n rangle n = alpha x - 2alpha langle x, n rangle n = alpha R(x) .$$
It therefore suffices to show $lVert R(x) rVert =lVert x rVert$.
Simplifying your computations we get
$$lVert R(x) rVert^2 = langle R(x), R(x) rangle = langle x - 2langle x, n rangle n, x - 2langle x, n rangle n rangle \ = langle x, x rangle - 2langle x, n rangle langle n, x rangle - 2langle x, n rangle langle x, n rangle + 4langle x, n rangle^2 langle n, n rangle \ = lVert x rVert^2 - 4 langle x, n rangle ^2 + 4 langle x, n rangle ^2 lVert n rVert^2 .$$
You did not mention the assumption $lVert n rVert = 1$. But in that case you see that
$$lVert R(x) rVert^2 = lVert x rVert^2 .$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sir I have put hat on $n$ represents that $n$ is unit vector! SIr Thank You so much
    $endgroup$
    – Noor Aslam
    Jan 17 at 15:17










  • $begingroup$
    I see! But for future questions I recommend to explain precisely everything concerning notation. Otherwise you are in danger that some readers will not understand it correctly.
    $endgroup$
    – Paul Frost
    Jan 17 at 16:08
















0












$begingroup$

Let us write $x$ instead of $vec{x}$ and $n$ instead of $hat{n}$. Moreover, $bullet$ denotes the usual dot product (scalar product) on Euclidean space $mathbb{R}^3$. In other words, $a bullet b = langle a, b rangle$ which is the alternative notation used in your question.



$R$ is a linear map because
$$R(x + y) = x + y - 2langle x + y, n rangle n = x + y - 2(langle x, n rangle + langle y, n rangle )n \= x - 2langle x, n rangle n + y - 2langle y, n rangle n = R(x) + R(y)$$
and
$$R(alpha x) = alpha x - 2langle alpha x, n rangle n = alpha x - 2alpha langle x, n rangle n = alpha R(x) .$$
It therefore suffices to show $lVert R(x) rVert =lVert x rVert$.
Simplifying your computations we get
$$lVert R(x) rVert^2 = langle R(x), R(x) rangle = langle x - 2langle x, n rangle n, x - 2langle x, n rangle n rangle \ = langle x, x rangle - 2langle x, n rangle langle n, x rangle - 2langle x, n rangle langle x, n rangle + 4langle x, n rangle^2 langle n, n rangle \ = lVert x rVert^2 - 4 langle x, n rangle ^2 + 4 langle x, n rangle ^2 lVert n rVert^2 .$$
You did not mention the assumption $lVert n rVert = 1$. But in that case you see that
$$lVert R(x) rVert^2 = lVert x rVert^2 .$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sir I have put hat on $n$ represents that $n$ is unit vector! SIr Thank You so much
    $endgroup$
    – Noor Aslam
    Jan 17 at 15:17










  • $begingroup$
    I see! But for future questions I recommend to explain precisely everything concerning notation. Otherwise you are in danger that some readers will not understand it correctly.
    $endgroup$
    – Paul Frost
    Jan 17 at 16:08














0












0








0





$begingroup$

Let us write $x$ instead of $vec{x}$ and $n$ instead of $hat{n}$. Moreover, $bullet$ denotes the usual dot product (scalar product) on Euclidean space $mathbb{R}^3$. In other words, $a bullet b = langle a, b rangle$ which is the alternative notation used in your question.



$R$ is a linear map because
$$R(x + y) = x + y - 2langle x + y, n rangle n = x + y - 2(langle x, n rangle + langle y, n rangle )n \= x - 2langle x, n rangle n + y - 2langle y, n rangle n = R(x) + R(y)$$
and
$$R(alpha x) = alpha x - 2langle alpha x, n rangle n = alpha x - 2alpha langle x, n rangle n = alpha R(x) .$$
It therefore suffices to show $lVert R(x) rVert =lVert x rVert$.
Simplifying your computations we get
$$lVert R(x) rVert^2 = langle R(x), R(x) rangle = langle x - 2langle x, n rangle n, x - 2langle x, n rangle n rangle \ = langle x, x rangle - 2langle x, n rangle langle n, x rangle - 2langle x, n rangle langle x, n rangle + 4langle x, n rangle^2 langle n, n rangle \ = lVert x rVert^2 - 4 langle x, n rangle ^2 + 4 langle x, n rangle ^2 lVert n rVert^2 .$$
You did not mention the assumption $lVert n rVert = 1$. But in that case you see that
$$lVert R(x) rVert^2 = lVert x rVert^2 .$$






share|cite|improve this answer









$endgroup$



Let us write $x$ instead of $vec{x}$ and $n$ instead of $hat{n}$. Moreover, $bullet$ denotes the usual dot product (scalar product) on Euclidean space $mathbb{R}^3$. In other words, $a bullet b = langle a, b rangle$ which is the alternative notation used in your question.



$R$ is a linear map because
$$R(x + y) = x + y - 2langle x + y, n rangle n = x + y - 2(langle x, n rangle + langle y, n rangle )n \= x - 2langle x, n rangle n + y - 2langle y, n rangle n = R(x) + R(y)$$
and
$$R(alpha x) = alpha x - 2langle alpha x, n rangle n = alpha x - 2alpha langle x, n rangle n = alpha R(x) .$$
It therefore suffices to show $lVert R(x) rVert =lVert x rVert$.
Simplifying your computations we get
$$lVert R(x) rVert^2 = langle R(x), R(x) rangle = langle x - 2langle x, n rangle n, x - 2langle x, n rangle n rangle \ = langle x, x rangle - 2langle x, n rangle langle n, x rangle - 2langle x, n rangle langle x, n rangle + 4langle x, n rangle^2 langle n, n rangle \ = lVert x rVert^2 - 4 langle x, n rangle ^2 + 4 langle x, n rangle ^2 lVert n rVert^2 .$$
You did not mention the assumption $lVert n rVert = 1$. But in that case you see that
$$lVert R(x) rVert^2 = lVert x rVert^2 .$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 14:25









Paul FrostPaul Frost

12.5k31035




12.5k31035












  • $begingroup$
    Sir I have put hat on $n$ represents that $n$ is unit vector! SIr Thank You so much
    $endgroup$
    – Noor Aslam
    Jan 17 at 15:17










  • $begingroup$
    I see! But for future questions I recommend to explain precisely everything concerning notation. Otherwise you are in danger that some readers will not understand it correctly.
    $endgroup$
    – Paul Frost
    Jan 17 at 16:08


















  • $begingroup$
    Sir I have put hat on $n$ represents that $n$ is unit vector! SIr Thank You so much
    $endgroup$
    – Noor Aslam
    Jan 17 at 15:17










  • $begingroup$
    I see! But for future questions I recommend to explain precisely everything concerning notation. Otherwise you are in danger that some readers will not understand it correctly.
    $endgroup$
    – Paul Frost
    Jan 17 at 16:08
















$begingroup$
Sir I have put hat on $n$ represents that $n$ is unit vector! SIr Thank You so much
$endgroup$
– Noor Aslam
Jan 17 at 15:17




$begingroup$
Sir I have put hat on $n$ represents that $n$ is unit vector! SIr Thank You so much
$endgroup$
– Noor Aslam
Jan 17 at 15:17












$begingroup$
I see! But for future questions I recommend to explain precisely everything concerning notation. Otherwise you are in danger that some readers will not understand it correctly.
$endgroup$
– Paul Frost
Jan 17 at 16:08




$begingroup$
I see! But for future questions I recommend to explain precisely everything concerning notation. Otherwise you are in danger that some readers will not understand it correctly.
$endgroup$
– Paul Frost
Jan 17 at 16:08


















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