Forming an orthonormal vector when you already have two perpendicular vectors

So I understand the requirements for an orthonormal basis and everything around it. However, there's one thing I am missing:

Suppose you have two vectors which are orthonormal $u_1$ and $u_2$. According to the answerbook the multiplication of vector $u_1$ and $u_2$ results in another orthonormal vector $u_3$.

Is this an actual standard theory? Does the multiplication of two orthonormal vectors results in another orthonormal vector?

I have included a picture just to make it more clear.

Thank you in advance :)

• Yes, for two vectors in $\mathbb R^{3}$ which are orthonormal, the cross product always gives you a third vector making the three vectors orthonormal. – Kavi Rama Murthy Sep 14 '18 at 9:49
• You do want to be careful: "multiplication of vectors" could refer to the dot product or the cross product. Since you're talking about the result being a vector, this implies you mean the cross product (also, we know the dot product of orthogonal vectors more readily than their cross product). – Teepeemm Sep 14 '18 at 14:26

Here we are using the property of cross product which is defined only for $v\in \mathbb{R^3}$.

The method is therefore not useful in general but it is very effective in that case to find an orthonormal basis.

The cross-product $u_1\times u_2$ of any two vectors $u_1$ and $u_2$ is always orthogonal to both of them. Furthermore,$$\lVert u_1\times u_2\rVert=\lVert u_1\rVert.\lVert u_2\rVert.\sin\theta,$$where $\theta$ is the angle between them. Therefore, if $u_1$ and $u_2$ are orthogonal and both of them have norm $1$, $u_1\times u_2$ will also have norm $1$ (and it will be orthogonal to the other two).

Yes, you can check it by direct calculation.

Assume that $u = \begin{bmatrix} u_1 \\ u_2 \\ u_3 \end{bmatrix}$ and $v = \begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix}$ are orthonormal vectors. Their cross product is defined as $$u \times v = \begin{bmatrix} u_2v_3 - u_3v_2 \\ u_3v_1-u_1v_3 \\ u_1v_2 - u_2v_1 \end{bmatrix}$$

We have

$$\langle u, u \times v\rangle = u_1u_2v_3 - u_1u_3v_2 + u_2u_3v_1 - u_1u_2v_3 + u_1u_3v_2 - u_2u_3v_1 = 0$$

$$\langle v, u \times v\rangle = u_2v_1v_3 - u_3v_1v_2 + u_2v_1v_2 - u_1v_2v_3 + u_1v_2v_3 - u_2v_1v_3 = 0$$

\begin{align} \|u\times v\|^2 &= (u_2v_3 - u_3v_2)^2 + (u_3v_1-u_1v_3)^2 + (u_1v_2 - u_2v_1)^2 \\ &= u_2^2v_3^2 - 2u_2u_3v_2v_3 + u_3^2v_2^2 + u_3^2v_1^2 - 2u_1u_3v_1v_3 + u_1^2v_3^2 + u_1^2v_2^2 - 2u_1u_2v_1v_2 + u_2^2v_1^2 \\ &= u_1^2(v_2^2 + v_3^2) + u_2^2(v_1^2 + v_3^2) + u_3^2(v_1^2 + v_2^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \\ &= u_1^2(1-v_1^2) + u_2^2(1 -v_2^2) + u_3^2(1-v_3^2) - 2u_2u_3v_2v_3 - 2u_1u_3v_1v_3 - 2u_1u_2v_1v_2 \\ &= (u_1^2+u_2^2+u_3^2) - (u_1^2v_1^2 + u_2^2v_2^2+u_3^2v_3^2 + 2u_1u_2v_1v_2 + 2u_1u_3v_1v_3 + 2u_2u_3v_2v_3)\\ &= \|u\|^2 - \langle u,v\rangle ^2\\ &= 1 \end{align}

so $\{u, v, u\times v\}$ is an orthonormal basis for $\mathbb{R}^3$.