An orthonormal set of vectors is a special collection where every vector satisfies two crucial properties:

• Orthogonal: Any two distinct vectors within the set are perpendicular to each other. Mathematically, their dot product is zero.
• Normalized: Each vector in the set has a magnitude (or length) of exactly 1. These are also known as unit vectors.

In essence, orthonormal vectors are both orthogonal (at right angles) and normalized (of unit length). This combination makes them incredibly useful as a basis for vector spaces because they are independent and have a consistent scale.

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:

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