How to show isometry of the space through plane?

Question

I am totally new to the isometries of the plane and space. I have to prove that the map $$R(\vec{x})=\vec{x}-2(\vec{x}\bullet\hat{n})\hat{n}$$ 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 $$R(\vec{x_{1}})=\vec{x_{1}}-2(\vec{x_{1}}\bullet\hat{n})\hat{n}$$ and $$R(\vec{x_{2}})=\vec{x_{2}}-2(\vec{x_{2}}\bullet\hat{n})\hat{n}$$ 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 $R(\vec{x_{1}})-R(\vec{x_{2}})=(\vec{x_{1}}-\vec{x_{2}})-2\big((\vec{x_{1}}-\vec{x_{2}})\bullet\hat{n}\big)\hat{n}$ Now $$||R(\vec{x_{1}})-R(\vec{x_{2}})||^{2}=\big<(\vec{x_{1}}-\vec{x_{2}})-2\big((\vec{x_{1}}-\vec{x_{2}})\bullet\hat{n}\big)\hat{n}, (\vec{x_{1}}-\vec{x_{2}})-2\big((\vec{x_{1}}-\vec{x_{2}})\bullet\hat{n}\big)\hat{n}\big>$$ I get confused, how to proceed step by step to reach that $$\big<(\vec{x_{1}}-\vec{x_{2}})-2\big((\vec{x_{1}}-\vec{x_{2}})\bullet\hat{n}\big)\hat{n}, (\vec{x_{1}}-\vec{x_{2}})-2\big((\vec{x_{1}}-\vec{x_{2}})\bullet\hat{n}\big)\hat{n}\big>=||\vec{x_{1}}-\vec{x_{2}}||^{2}$$ $$\Rightarrow~ \bigg<\vec{x_{1}}-\vec{x_{2}},\vec{x_{1}}-\vec{x_{2}}\bigg>-4\big[(\vec{x_{1}}-\vec{x_{2}})\bullet\hat{n}\big]\bigg<\vec{x_{1}}-\vec{x_{2}},\hat{n}\bigg>+4\bigg(\big[(\vec{x_{1}}-\vec{x_{2}})\bullet\hat{n}\big]\bigg)^{2}\bigg<\hat{n},\hat{n}\bigg>$$ Am I right, how to show that $$4\big[(\vec{x_{1}}-\vec{x_{2}})\bullet\hat{n}\big]\bigg<\vec{x_{1}}-\vec{x_{2}},\hat{n}\bigg>=4\bigg(\big[(\vec{x_{1}}-\vec{x_{2}})\bullet\hat{n}\big]\bigg)^{2}$$

Answer 1

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