Question

Find two unit vectors orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\) if: $$ \begin{array}{l} \text { a. } \mathbf{u}=\left[\begin{array}{l} 1 \\ 2 \\ 2 \end{array}\right], \mathbf{v}=\left[\begin{array}{r} 2 \\ -1 \\ 2 \end{array}\right] \\ \text { b. } \mathbf{u}=\left[\begin{array}{r} 1 \\ 2 \\ -1 \end{array}\right], \mathbf{v}=\left[\begin{array}{l} 3 \\ 1 \\ 2 \end{array}\right] \end{array} $$

Short answer

The unit vectors orthogonal to \(\mathbf{u}\) and \(\mathbf{v}\) for a) are \(\begin{bmatrix} \frac{6}{\sqrt{65}} \\ \frac{-2}{\sqrt{65}} \\ \frac{-5}{\sqrt{65}} \end{bmatrix}\) and its negative, and for b) are \(\begin{bmatrix} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ -\frac{1}{\sqrt{3}} \end{bmatrix}\) and its negative.

Step-by-step solution

Calculate the Cross Product for Part (a)

Find the cross product \( \mathbf{w} = \mathbf{u} \times \mathbf{v} \) for the vectors \( \mathbf{u} = \begin{bmatrix} 1 \ 2 \ 2 \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} 2 \ -1 \ 2 \end{bmatrix} \). The cross product is calculated as follows:\[ \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & 2 \ 2 & -1 & 2 \end{vmatrix} = \mathbf{i}(2 \cdot 2 - (-1) \cdot 2) - \mathbf{j}(1 \cdot 2 - 2 \cdot 2) + \mathbf{k}(1 \cdot (-1) - 2 \cdot 2) \]\[ \mathbf{w} = \begin{bmatrix} 6 \ -2 \ -5 \end{bmatrix} \]

Normalize the Cross Product for Part (a)

Normalize the vector \( \mathbf{w} = \begin{bmatrix} 6 \ -2 \ -5 \end{bmatrix} \) to find a unit vector. First, compute its magnitude:\[ \|\mathbf{w}\| = \sqrt{6^2 + (-2)^2 + (-5)^2} = \sqrt{36 + 4 + 25} = \sqrt{65} \]The unit vector in the direction of \( \mathbf{w} \) is:\[ \mathbf{w}_{unit} = \frac{1}{\sqrt{65}} \begin{bmatrix} 6 \ -2 \ -5 \end{bmatrix} = \begin{bmatrix} \frac{6}{\sqrt{65}} \ \frac{-2}{\sqrt{65}} \ \frac{-5}{\sqrt{65}} \end{bmatrix} \]

Find the Second Orthogonal Unit Vector for Part (a)

Since the cross product \( \mathbf{w} \) is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \), the second orthogonal direction is simply the negative of \( \mathbf{w}_{unit} \):\[ -\mathbf{w}_{unit} = \begin{bmatrix} -\frac{6}{\sqrt{65}} \ \frac{2}{\sqrt{65}} \ \frac{5}{\sqrt{65}} \end{bmatrix} \]

Calculate the Cross Product for Part (b)

Find the cross product \( \mathbf{w} = \mathbf{u} \times \mathbf{v} \) for the vectors \( \mathbf{u} = \begin{bmatrix} 1 \ 2 \ -1 \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} 3 \ 1 \ 2 \end{bmatrix} \). The cross product is calculated as follows:\[ \mathbf{w} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 2 & -1 \ 3 & 1 & 2 \end{vmatrix} = \mathbf{i}(2 \cdot 2 - (-1) \cdot 1) - \mathbf{j}(1 \cdot 2 - 3 \cdot (-1)) + \mathbf{k}(1 \cdot 1 - 2 \cdot 3) \]\[ \mathbf{w} = \begin{bmatrix} 5 \ 5 \ -5 \end{bmatrix} \]

Normalize the Cross Product for Part (b)

Normalize the vector \( \mathbf{w} = \begin{bmatrix} 5 \ 5 \ -5 \end{bmatrix} \) to find a unit vector. First, compute its magnitude:\[ \|\mathbf{w}\| = \sqrt{5^2 + 5^2 + (-5)^2} = \sqrt{75} = 5\sqrt{3} \]The unit vector in the direction of \( \mathbf{w} \) is:\[ \mathbf{w}_{unit} = \frac{1}{5\sqrt{3}} \begin{bmatrix} 5 \ 5 \ -5 \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{3}} \ \frac{1}{\sqrt{3}} \ -\frac{1}{\sqrt{3}} \end{bmatrix} \]

Find the Second Orthogonal Unit Vector for Part (b)

Since the cross product \( \mathbf{w} \) is perpendicular to both \( \mathbf{u} \) and \( \mathbf{v} \), the second orthogonal direction is simply the negative of \( \mathbf{w}_{unit} \):\[ -\mathbf{w}_{unit} = \begin{bmatrix} -\frac{1}{\sqrt{3}} \ -\frac{1}{\sqrt{3}} \ \frac{1}{\sqrt{3}} \end{bmatrix} \]

Key concepts

Cross Product

The cross product is a vector operation used to find a vector that is orthogonal, or perpendicular, to two other vectors in three-dimensional space. When you have two vectors, let's say \( \mathbf{u} = \begin{bmatrix} 1 \ 2 \ 2 \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} 2 \ -1 \ 2 \end{bmatrix} \) as given in the exercise, the cross product \( \mathbf{w} = \mathbf{u} \times \mathbf{v} \) is calculated using the determinant of a matrix, which involves the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \). This results in another vector \( \mathbf{w} \) that is orthogonal to both \( \mathbf{u} \) and \( \mathbf{v} \). For example, the calculation yields the vector \( \begin{bmatrix} 6 \ -2 \ -5 \end{bmatrix} \) which is perpendicular to both original vectors. This technique is particularly useful in physics and engineering for finding normal vectors to planes defined by two vectors.

Unit Vectors

Unit vectors are vectors with a magnitude, or length, of exactly 1. They are often used to indicate direction and remove any scaling information from a vector. To obtain a unit vector from any given vector, you divide the vector by its magnitude. This process keeps the vector pointing in the same direction, but scales its length to 1. In the exercise, after finding a cross product vector \( \mathbf{w} \), it's converted to a unit vector so it retains the orthogonal property but with a length of 1. For instance, the unit vector of \( \mathbf{w} = \begin{bmatrix} 6 \ -2 \ -5 \end{bmatrix} \) becomes \( \mathbf{w}_{unit} = \begin{bmatrix} \frac{6}{\sqrt{65}} \ \frac{-2}{\sqrt{65}} \ \frac{-5}{\sqrt{65}} \end{bmatrix} \). Using unit vectors is crucial in simplifying vector math and ensuring precision in directionality.

Vector Normalization

Vector normalization is the process of converting a vector into a unit vector that points in the same direction as the original. This is achieved by dividing each component of the vector by its magnitude. The purpose of normalization is to maintain the direction of the original vector while standardizing its length to 1, hence making it a unit vector. For the vector \( \mathbf{w} = \begin{bmatrix} 6 \ -2 \ -5 \end{bmatrix} \), normalization involves calculating its magnitude \( \|\mathbf{w}\| = \sqrt{6^2 + (-2)^2 + (-5)^2} = \sqrt{65} \), and then the normalized vector becomes \( \begin{bmatrix} \frac{6}{\sqrt{65}} \ \frac{-2}{\sqrt{65}} \ \frac{-5}{\sqrt{65}} \end{bmatrix} \). This step is essential because it allows vectors to be compared and combined more easily while preserving their directional properties.

Magnitude Calculation

Magnitude, in vector mathematics, is the length of a vector. It is the measure of how long the vector is, irrespective of the direction it points. To find the magnitude, also known as the norm, of a vector \( \mathbf{w} = \begin{bmatrix} x \ y \ z \end{bmatrix} \), you apply the formula \( \|\mathbf{w}\| = \sqrt{x^2 + y^2 + z^2} \). In the provided exercise, the vector \( \mathbf{w} = \begin{bmatrix} 6 \ -2 \ -5 \end{bmatrix} \) has a magnitude calculated as \( \sqrt{65} \). This calculation is crucial because it provides the scale of a vector, facilitating conversions to unit vectors via normalization. Understanding and applying magnitude is key in numerous fields, including physics, where it helps quantify forces, and 3D graphics, where it helps in scaling and transformations.