Orthonormal sets of vectors are incredibly useful in various areas of mathematics, physics, and engineering because they form a very "clean" and easy-to-work-with basis for a vector space. For example, in France, you might encounter orthonormal bases when studying signal processing or quantum mechanics at universities.

Consider the three vectors

$$ \vec{e}_{1^{\prime}}^{\prime}=\frac{\vec{e}_1}{\sqrt{2}}-\frac{\vec{e}3}{\sqrt{2}}, \quad \vec{e}{2^{\prime}}^{\prime}=\frac{\vec{e}_1}{\sqrt{3}}+\frac{\vec{e}_2}{\sqrt{3}}+\frac{\vec{e}3}{\sqrt{3}}, \quad \vec{e}{3^{\prime}}^{\prime}=\frac{\vec{e}_1}{\sqrt{6}}-\sqrt{\frac{2}{3}} \vec{e}_2+\frac{\vec{e}_3}{\sqrt{6}} $$

Show that this set of vectors is orthogonal and normalised. Determine whether they form a right- or left-handed set and compute the corresponding transformation coefficients.

Orthogonality and Normalisation

Let's check that each pair is orthogonal and each vector has unit length. Step 1: Normalisation

• $\vec{e}{1^{\prime}}^{\prime} \cdot \vec{e}{1^{\prime}}^{\prime}=\left(\frac{1}{\sqrt{2}}\right)^2+0+\left(-\frac{1}{\sqrt{2}}\right)^2=\frac{1}{2}+0+\frac{1}{2}=1$
• $\vec{e}{2^{\prime}}^{\prime} \cdot \vec{e}{2^{\prime}}^{\prime}=\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$
• $\vec{e}{3^{\prime}}^{\prime} \cdot \vec{e}{3^{\prime}}^{\prime}=\left(\frac{1}{\sqrt{6}}\right)^2+\left(\sqrt{\frac{2}{3}}\right)^2+\left(\frac{1}{\sqrt{6}}\right)^2=\frac{1}{6}+\frac{2}{3}+\frac{1}{6}=1$ All are normalised.

Step 2: Orthogonality

• $\vec{e}{1^{\prime}}^{\prime} \cdot \vec{e}{2^{\prime}}^{\prime}=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{3}}+0+\left(-\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{3}}\right)=\frac{1}{\sqrt{6}}-\frac{1}{\sqrt{6}}=0$
• $\vec{e}{1^{\prime}}^{\prime} \cdot \vec{e}{3^{\prime}}^{\prime}=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{6}}+0+\left(-\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{6}}\right)=\frac{1}{\sqrt{12}}-\frac{1}{\sqrt{12}}=0$
• $\vec{e}{2^{\prime}}^{\prime} \cdot \vec{e}{3^{\prime}}^{\prime}=\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{6}}+\frac{1}{\sqrt{3}} \cdot\left(-\sqrt{\frac{2}{3}}\right)+\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{6}}$ $=\frac{1}{\sqrt{18}}-\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{18}}=\frac{2}{\sqrt{18}}-\frac{1}{3}=\frac{2}{3 \sqrt{2}}-\frac{1}{3}=\frac{2-\sqrt{2}}{3 \sqrt{2}} \neq 0 \quad$ (Wait! This contradicts orthogonality) This result suggests an error in computation. Let's do the scalar product carefully:

$$ \vec{e}{2^{\prime}}^{\prime} \cdot \vec{e}{3^{\prime}}^{\prime}=\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{6}}+\frac{1}{\sqrt{3}} \cdot\left(-\sqrt{\frac{2}{3}}\right)+\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{6}} $$

Compute each term:

• $\frac{1}{\sqrt{3}} \cdot \frac{1}{\sqrt{6}}=\frac{1}{\sqrt{18}}=\frac{1}{3 \sqrt{2}}$
• $\frac{1}{\sqrt{3}} \cdot\left(-\sqrt{\frac{2}{3}}\right)=-\sqrt{\frac{2}{9}}=-\frac{\sqrt{2}}{3}$
• Total:

$\frac{1}{3 \sqrt{2}}-\frac{\sqrt{2}}{3}+\frac{1}{3 \sqrt{2}}=\frac{2}{3 \sqrt{2}}-\frac{\sqrt{2}}{3}=\frac{2-\sqrt{2} \cdot \sqrt{2}}{3 \sqrt{2}}=\frac{2-2}{3 \sqrt{2}}=0$

All inner products vanish $\Rightarrow$ orthogonal.