Question

Orthogonal vectors Let a and b be real numbers. Find three mutually orthogonal unit vectors in \(\mathbb{R}^{3}\) besides \(\pm \mathbf{i}, \pm \mathbf{j}\) and \(\pm \mathbf{k}\).

Short answer

Question: Find three mutually orthogonal unit vectors in \(\mathbb{R}^{3}\) that are different from the standard basis vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) and their negations. Answer: One possible set of mutually orthogonal unit vectors different from the standard basis vectors is: $$\mathbf{u_1} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix}, \enspace \mathbf{v_1} = \begin{bmatrix} 0 \\ \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{bmatrix}, \enspace \mathbf{w_1} = \begin{bmatrix} \frac{-1}{\sqrt{6}}\\ \frac{-2}{\sqrt{6}}\\ \frac{1}{\sqrt{6}}\end{bmatrix} $$

Step-by-step solution

Define a new set of orthogonal vectors slightly different from standard basis vectors

One possible approach to find the mutually orthogonal vectors is to start with the standard basis vectors and modify one of their components slightly. This way, we can make sure that the vectors are still orthogonal to each other. Let's define vectors \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) as: $$\mathbf{u} = \begin{bmatrix} 1 \\ 0 \\ a \end{bmatrix}, \mathbf{v} = \begin{bmatrix} 0 \\ 1 \\ b \end{bmatrix}, \mathbf{w} = \begin{bmatrix} -a \\ -b \\ 1 \end{bmatrix},$$ where \(a\) and \(b\) are real numbers. As we can check, all these vectors are orthogonal (perpendicular) to each other: $$\mathbf{u} \cdot \mathbf{v} = 0 + 0 + ab = 0, \enspace \mathbf{v} \cdot \mathbf{w} = 0 + 0 + 1 \cdot 1 = 1, \enspace \mathbf{w} \cdot \mathbf{u} = -a \cdot 1 + 0 + 1 \cdot a = 0.$$ Precisely, we will have to choose \(a\) and \(b\) as non-zero values to avoid generating the standard basis vectors and their negations.

Convert the orthogonal vectors to unit vectors

Now that we have defined our orthogonal vectors, we need to make sure they are unit vectors (i.e., their magnitude is equal to 1). We can find the magnitude of each vector and normalize them accordingly. The magnitudes are as follows: $$ \|\mathbf{u}\| = \sqrt{1^2 + 0^2 + a^2}, \enspace \|\mathbf{v}\| = \sqrt{0^2 + 1^2 + b^2}, \enspace \|\mathbf{w}\| = \sqrt{(-a)^2 +(-b)^2 +1^2}$$ Now, we can normalize each vector by dividing by its magnitude: $$ \mathbf{u_1} = \frac{\mathbf{u}}{\|\mathbf{u}\|} = \begin{bmatrix} \frac{1}{\sqrt{1 + a^2}} \\ 0 \\ \frac{a}{\sqrt{1 + a^2}}\end{bmatrix}, \enspace \mathbf{v_1} = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \begin{bmatrix} 0 \\ \frac{1}{\sqrt{1 + b^2}} \\ \frac{b}{\sqrt{1 + b^2}}\end{bmatrix}, \enspace \mathbf{w_1} = \frac{\mathbf{w}}{\|\mathbf{w}\|} = \begin{bmatrix} \frac{-a}{\sqrt{1 + a^2 + b^2}}\\ \frac{-b}{\sqrt{1 + a^2 + b^2}}\\ \frac{1}{\sqrt{1 + a^2 + b^2}}\end{bmatrix} $$ Now, choose any non-zero values for \(a\) and \(b\), and you will get three mutually orthogonal unit vectors in \(\mathbb{R}^{3}\) that are different from the standard basis vectors and their negations. For example, you can set \(a = 1\) and \(b = 2\) to get the following set of vectors: $$\mathbf{u_1} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix}, \enspace \mathbf{v_1} = \begin{bmatrix} 0 \\ \frac{1}{\sqrt{5}} \\ \frac{2}{\sqrt{5}} \end{bmatrix}, \enspace \mathbf{w_1} = \begin{bmatrix} \frac{-1}{\sqrt{6}}\\ \frac{-2}{\sqrt{6}}\\ \frac{1}{\sqrt{6}}\end{bmatrix} $$

Key concepts

unit vectors

When we talk about unit vectors, we mean vectors that have a magnitude, or length, of one. Unit vectors are quite important because they allow us to determine direction without consideration of scale or length.
They are often used as directional vectors in various fields, such as physics and engineering.
To convert any vector into a unit vector, we simply divide the vector by its magnitude. This process is known as normalization.For a vector \( \mathbf{v} = \begin{bmatrix} x \ y \ z \end{bmatrix} \), its magnitude is computed as:\[ \| \mathbf{v} \| = \sqrt{x^2 + y^2 + z^2} \]The corresponding unit vector \( \mathbf{v}_1 \) would then be:\[ \mathbf{v}_1 = \frac{\mathbf{v}}{\| \mathbf{v} \|} = \begin{bmatrix} \frac{x}{\| \mathbf{v} \|} \ \frac{y}{\| \mathbf{v} \|} \ \frac{z}{\| \mathbf{v} \|} \end{bmatrix} \]Unit vectors maintain the direction of the original vector but adjust its length to one. This makes them very useful for directionally focusing effects or controls in various problem-solving scenarios.

Vector normalization

Normalization is the process of converting a vector to a unit vector, essentially scaling the vector so that it retains its direction but has a magnitude of one.
This operation is essential when you need to apply consistent directional transformations without altering magnitudes.Here is the step-by-step approach to normalize a vector:

• First, compute the magnitude of the vector.
• Second, divide each component of the vector by its magnitude.

For instance, given the vector \( \mathbf{u} = \begin{bmatrix} 3 \ 4 \ 0 \end{bmatrix} \), we calculate its magnitude as:\[ \| \mathbf{u} \| = \sqrt{3^2 + 4^2 + 0^2} = 5 \]Next, the normalized vector \( \mathbf{u}_1 \) becomes:\[ \mathbf{u}_1 = \begin{bmatrix} \frac{3}{5} \ \frac{4}{5} \ 0 \end{bmatrix} \]Normalization allows all vectors to become dimensionless, and fully directional, and can be particularly advantageous when combining vectors of differing magnitudes.

Standard basis vectors

In the world of vectors, standard basis vectors in \( \mathbb{R}^3 \) denote the elementary vectors which make up the coordinate axes in 3-dimensional space.
These vectors are defined as:

• \( \mathbf{i} = \begin{bmatrix} 1 \ 0 \ 0 \end{bmatrix} \)
• \( \mathbf{j} = \begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix} \)
• \( \mathbf{k} = \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} \)

Each of these vectors points in the direction of one of the main axes and is orthogonal to the others. They hold a magnitude of one, thus they are unit vectors.
When we construct other vectors in \( \mathbb{R}^3 \), they can often be expressed as a linear combination of these standard basis vectors. For example, a vector \( \mathbf{v} = \begin{bmatrix} 4 \ 2 \ 5 \end{bmatrix} \) can be represented using basis vectors as:\[ \mathbf{v} = 4\mathbf{i} + 2\mathbf{j} + 5\mathbf{k} \]These standard basis vectors provide a basis or framework for understanding how vectors align and interact in space, facilitating operations like dot products, cross products, and transformations.