For orthonormal vectors $a, b \in \mathbb{R}^3$ prove $\det\begin{bmatrix} a & b & a \times b \end{bmatrix} = 1$.

Question

Let $a = (a_1, a_2, a_3)$ and $b = (b_1, b_2, b_3)$ be two orthonormal vectors in $\mathbb{R}^3$ and $a \times b \in \mathbb{R}^3$ their cross product.

I would like to prove by direct calculation that the matrix $A = \begin{bmatrix} a & b & a \times b \end{bmatrix}$ having the vectors $a, b$ and $a \times b$ as its columns has determinant $1$.

We could for example observe that $A$ represents a transformation from an orthonormal basis $\{e_1, e_2, e_3\}$ to an orthonormal basis $\{a, b, a \times b\}$ which could be achieved as a composition of two rotations (one rotating $e_3$ to $a \times b$, and another around the axis $a \times b$ aligning $\{e_1, e_2\}$ with $\{a, b\}$), and hence has determinant $1$. I'm not interested in such approaches.

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My attempt:

We have $a\times b = \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_2b_2 - a_2b_1\end{pmatrix}$.

Thus: \begin{align} \det A &= \begin{vmatrix} a_1 & b_1 & a_2b_3 - a_3b_2 \\ a_2 & b_2 & a_3b_1 - a_1b_3 \\ a_3 & b_3 & a_1b_2 - a_2b_1 \\ \end{vmatrix} \\ &= \text{Laplace expansion along the first column}\\ &= a_1 \begin{vmatrix} b_2 & a_3b_1 - a_1b_3 \\ b_3 & a_1b_2 - a_2b_1 \\ \end{vmatrix} - a_2 \begin{vmatrix} b_1 & a_2b_3 - a_3b_2 \\ b_3 & a_1b_2 - a_2b_1 \\ \end{vmatrix} + a_3 \begin{vmatrix} b_1 & a_2b_3 - a_3b_2 \\ b_2 & a_3b_1 - a_1b_3 \\ \end{vmatrix} \\ &= a_1 \big(a_1{b_2}^2 - a_2b_1b_2 - a_3b_1b_3 + a_1{b_3}^2\big) - a_2\big(a_1b_1b_2 - a_2{b_1}^2 - a_2{b_3}^2 + a_3b_2b_3\big) + a_3\big(a_3{b_1}^2 - a_1b_1b_3 - a_2b_2b_3 + a_3{b_2}^2\big) \\ &= \color{red}{{a_1}^2{b_2}^2} - \color{green}{a_1a_2b_1b_2} - a_1a_3b_1b_3 + \color{red}{{a_1}^2{b_3}^2} - \color{green}{a_1a_2b_1b_2} + \color{blue}{{a_2}^2{b_1}^2} + \color{blue}{{a_2}^2{b_3}^2} - a_2a_3b_2b_3 + {a_3}^2{b_1}^2 - a_1a_3b_1b_3 - a_2a_3b_2b_3 + {a_3}^2{b_2}^2 \\ &= \color{red}{{a_1}^2({b_2}^2 + {b_3}^2)} - \color{green}{2a_1a_2b_1b_2} - 2a_1a_3b_1b_3 + \color{blue}{{a_2}^2({b_1}^2+{b_3}^2)} - 2a_2a_3b_2b_3 + {a_3}^2({b_1}^2 + {b_2}^2) \\ \end{align}

Now we could use ${b_1}^2 + {b_2}^2 + {b_3}^2 = 1$ but nothing seems to simplify from this point onward. We also have to use orthogonality at some point. How should we proceed?

Answer 1

$$a_1^2(b^2_2+b_3^2) = a_1^2(1-b_1^2) = a_1^2 - a_1^2b_1^2$$

Repeat with the other terms and an $a_1^2+a_2^2+a_3^2 = 1$ will drop out.

Now you have a $\mathbf a \cdot \mathbf b$ appearing.

Answer 2

After fixing the typo in the calculation noticed by @Fan, the statement easily follows.

Using ${b_1}^2 + {b_2}^2 + {b_3}^2 = 1$ we obtain:

\begin{align}\det A &= \color{red}{{a_1}^2({b_2}^2 + {b_3}^2)} - \color{green}{2a_1a_2b_1b_2} - 2a_1a_3b_1b_3 + \color{blue}{{a_2}^2({b_1}^2+{b_3}^2)} - 2a_2a_3b_2b_3 + {a_3}^2({b_1}^2 + {b_2}^2) \\ &= {a_1}^2(1 - {b_1}^2) + {a_2}^2(1 - {b_2}^2) + {a_2}^2(1 - {b_2}^2) - 2a_1a_3b_1b_3 - 2a_2a_3b_2b_3 - 2a_1a_2b_1b_2\\ &= {a_1}^2 + {a_2}^2 + {a_3}^2 + \big({a_1}^2{b_1}^2 + {a_2}^2{b_2}^2 + {a_3}^2{b_3}^2 -2a_1a_3b_1b_3 - 2a_2a_3b_2b_3 - 2a_1a_2b_1b_2\big)\\ &= \|a\|^2 + \langle a, b\rangle^2\\ &= 1 \end{align}