Question

Find three mutually orthogonal unit vectors in \(\mathbb{R}^{3}\) besides \(\pm \mathbf{i}, \pm \mathbf{j},\) and \(\pm \mathbf{k}.\)

Short answer

Three mutually orthogonal unit vectors different from the standard unit vectors and their corresponding negatives are: $$ \mathbf{v}_1 = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0) \\ \mathbf{u} = (1, 1, 0) \\ \mathbf{v}_2' = (0, 0, -1) $$

Step-by-step solution

Find the first orthogonal vector

Choose an arbitrary vector that is not parallel to unit vector \(\mathbf{i}\). Let's consider \(\mathbf{u} = (1, 1, 0)\). Now, we will find a vector \(\mathbf{v}\) that is orthogonal to \(\mathbf{u}\). We know that for two vectors to be orthogonal, their dot product should be equal to 0. Therefore: $$ \mathbf{u} \cdot \mathbf{v} = 0 \implies \mathbf{u} \cdot (x, y, z) = (1, 1, 0) \cdot (x, y, z) = x + y = 0 $$ We can choose \(\mathbf{v}\) to be \((1, -1, 0)\), since it satisfies the equation above.

Normalize the first orthogonal vector

To make the vector \(\mathbf{v}\) from Step 1 a unit vector, we need to divide it by its magnitude. The magnitude of \(\mathbf{v}\) is: $$ \|\mathbf{v}\| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{2} $$ So, the first orthogonal unit vector \(\mathbf{v}_1\) is: $$ \mathbf{v}_1 = \frac{\mathbf{v}}{\|\mathbf{v}\|} = \frac{1}{\sqrt{2}}(1, -1, 0) = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0) $$

Use cross product to find the second orthogonal unit vector

We can use the cross product of \(\mathbf{u}\) and \(\mathbf{v}_1\) to find the second orthogonal unit vector \(\mathbf{v}_2\). The cross product \(\mathbf{u} \times \mathbf{v}_1\) is: $$ \mathbf{v}_2 =\mathbf{u} \times \mathbf{v}_1 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \end{vmatrix} = \mathbf{k} $$ Note that \(\mathbf{v}_2 = \mathbf{k}\), but since we are not allowed to use \(\mathbf{k}\), we can use the negatives of \(\mathbf{u}\) and \(\mathbf{v}_1\) instead.

Utilize negative vectors to find the revised second orthogonal unit vector

Use the cross product of \(-\mathbf{u}\) and \(-\mathbf{v}_1\) to find an alternate orthogonal unit vector to \(\mathbf{k}\). The cross product \((-\mathbf{u}) \times (-\mathbf{v}_1)\) is: $$ \mathbf{v}_2' = (-\mathbf{u}) \times (-\mathbf{v}_1) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & -1 & 0 \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{vmatrix} = -\mathbf{k} $$ So the second orthogonal unit vector is \(\mathbf{v}_2' = -\mathbf{k} = (0, 0, -1)\).

Combine the orthogonal unit vectors

We now have three mutually orthogonal unit vectors, which are different from the given vectors: $$ \mathbf{v}_1 = (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0) \\ \mathbf{u} = (1, 1, 0) \\ \mathbf{v}_2' = (0, 0, -1) $$

Key concepts

Unit Vectors

Unit vectors are essential in the field of mathematics and physics because they signify a vector with a magnitude of exactly one. Their primary role is direction indication without scaling the vector's length. In real-world applications, unit vectors often simplify complex calculations by setting a standard of length.
For a vector to be a unit vector, its magnitude, denoted as \( \|\mathbf{v}\| \), must equal 1. Mathematically, this is represented as:

• \( \| \mathbf{v} \| = 1 \)

Given a vector \( \mathbf{v} \), it can be transformed into a unit vector \( \mathbf{u} \) by dividing each component of \( \mathbf{v} \) by its magnitude:

• \( \mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|} \)

This normalization process ensures that the new vector \( \mathbf{u} \) maintains the original direction of \( \mathbf{v} \) while adopting a unified scale.

Dot Product

The dot product is a crucial operation for understanding vector relationships, particularly when examining orthogonality. This operation, also known as the scalar product, produces a scalar value that reflects the extent to which two vectors point in the same direction.
For vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the dot product is calculated as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)

If the result of this computation is zero, the vectors are orthogonal, meaning they form a right angle to each other. The concept of orthogonality is significant in various mathematical and engineering problems where independence or perpendicularity of directions is desired. Consequently, the dot product provides a straightforward means to determine such relationships.

Cross Product

Understanding vectors in three-dimensional space becomes more intuitive through the cross product. This operation results in a vector that is orthogonal to the two original vectors, which can be particularly useful in physics and engineering for finding perpendicular directions.
For two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the cross product \( \mathbf{a} \times \mathbf{b} \) is given by the determinant of a matrix formed with the standard unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \):

• \( \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \)

The result is a new vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), providing insight into the spatial orientation and interaction of vectors. In this way, the cross product is integral for conceptualizing three-dimensional vector relationships.

Magnitude

A vector's magnitude, or its length, is a fundamental concept when dealing with vector quantities. It represents how long a vector is, independent of its direction. Calculating the magnitude of a vector allows you to understand the scale of the force or movement it represents.
For a vector \( \mathbf{v} = (v_1, v_2, v_3) \), its magnitude is given by the formula:

• \( \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2} \)

This calculation uses the Pythagorean theorem to combine the vector's components into a single scalar value. Magnitude is especially important when normalizing a vector to transform it into a unit vector. It allows you to compare vector lengths and establish consistent measurements across different vectors in space, crucial for both theoretical and practical applications.