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Magazine Chapter 5: Problem 13
Determine whether the given orthogonal set of vectors is orthonormal. If it is not, normalize the vectors to form an orthonormal set. $$\left[\begin{array}{l}\frac{1}{3} \\\\\frac{2}{3} \\\\\frac{2}{3}\end{array}\right],\left[\begin{array}{r} \frac{2}{3} \\\\-\frac{1}{3} \\\0\end{array}\right],\left[\begin{array}{r}1 \\\2 \\\\-\frac{5}{2}\end{array}\right]$$
Short Answer
Expert verified
The set is not orthonormal; normalize vectors to form an orthonormal set.
Step by step solution
01
Check Orthogonality
To check if the vectors are orthonormal, we first confirm orthogonality by taking the dot product of each pair of vectors. The dot product for orthogonal vectors should be zero.
02
Compute Dot Products
Calculate the dot products: 1. \(\begin{bmatrix}\frac{1}{3} \\frac{2}{3} \\frac{2}{3}\end{bmatrix}\cdot\begin{bmatrix}\frac{2}{3} \-\frac{1}{3} \0\end{bmatrix} = \frac{1}{3}\cdot\frac{2}{3} + \frac{2}{3} \cdot -\frac{1}{3} + \frac{2}{3}\cdot0 = 0\)2. \(\begin{bmatrix}\frac{1}{3} \\frac{2}{3} \\frac{2}{3}\end{bmatrix}\cdot\begin{bmatrix}1 \2 \-\frac{5}{2}\end{bmatrix} = \frac{1}{3}\cdot1 + \frac{2}{3}\cdot2 + \frac{2}{3}\cdot-\frac{5}{2} = 0\)3. \(\begin{bmatrix}\frac{2}{3} \-\frac{1}{3} \0\end{bmatrix}\cdot\begin{bmatrix}1 \2 \-\frac{5}{2}\end{bmatrix} = \frac{2}{3}\cdot1 + -\frac{1}{3}\cdot2 + 0 = 0\)Since all dot products are zero, the vectors are mutually orthogonal.
03
Check Unit Length
Next, we check if each vector is of unit length by calculating the magnitude (or norm) of each vector and ensuring it equals one. The norm \( ||\mathbf{v}|| \) of vector \( \mathbf{v} = \begin{bmatrix} a \ b \ c \end{bmatrix} \) is \( \sqrt{a^2 + b^2 + c^2} \).
04
Calculate Norms
1. For \( \begin{bmatrix} \frac{1}{3} \ \frac{2}{3} \ \frac{2}{3} \end{bmatrix} \): \(||\mathbf{v}|| = \sqrt{ \left(\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^2} = \sqrt{\frac{1}{9} + \frac{4}{9} + \frac{4}{9}} = \sqrt{1} = 1\)2. For \( \begin{bmatrix} \frac{2}{3} \ -\frac{1}{3} \ 0 \end{bmatrix} \): \(||\mathbf{v}|| = \sqrt{\left(\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 + 0^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}\)3. For \( \begin{bmatrix} 1 \ 2 \ -\frac{5}{2} \end{bmatrix} \): \(||\mathbf{v}|| = \sqrt{1^2 + 2^2 + \left(-\frac{5}{2}\right)^2} = \sqrt{1 + 4 + \frac{25}{4}} = \sqrt{\frac{45}{4}} = \frac{3\sqrt{5}}{2}\)
05
Normalize Vectors
The second and third vectors are not unit length. Normalize them by dividing each component by their respective norms.1. Normalize \( \begin{bmatrix} \frac{2}{3} \ -\frac{1}{3} \ 0 \end{bmatrix} \): \[\frac{ \begin{bmatrix} \frac{2}{3} \ -\frac{1}{3} \ 0 \end{bmatrix} }{\frac{\sqrt{5}}{3}} = \begin{bmatrix} \frac{2}{\sqrt{5}} \ -\frac{1}{\sqrt{5}} \ 0 \end{bmatrix}\]2. Normalize \( \begin{bmatrix} 1 \ 2 \ -\frac{5}{2} \end{bmatrix} \): \[\frac{ \begin{bmatrix} 1 \ 2 \ -\frac{5}{2} \end{bmatrix} }{\frac{3\sqrt{5}}{2}} = \begin{bmatrix} \frac{2}{3\sqrt{5}} \ \frac{4}{3\sqrt{5}} \ -\frac{5}{3\sqrt{5}} \end{bmatrix}\]
06
Write Orthonormal Set
The orthonormal set of vectors is:1. \( \begin{bmatrix} \frac{1}{3} \ \frac{2}{3} \ \frac{2}{3} \end{bmatrix} \) (already unit length)2. \( \begin{bmatrix} \frac{2}{\sqrt{5}} \ -\frac{1}{\sqrt{5}} \ 0 \end{bmatrix} \)3. \( \begin{bmatrix} \frac{2}{3\sqrt{5}} \ \frac{4}{3\sqrt{5}} \ -\frac{5}{3\sqrt{5}} \end{bmatrix} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Orthogonal Vectors
Orthogonal vectors are pairs of vectors that, when "crossed" with each other via a mathematical operation called the dot product, result in zero. Imagine orthogonal vectors as being at a perfect right angle to each other in space. This implies they share no common direction.
To check for orthogonality in a group of vectors, compute the dot product between each pair. If all dot products equal zero, the vectors are orthogonal.
To check for orthogonality in a group of vectors, compute the dot product between each pair. If all dot products equal zero, the vectors are orthogonal.
- For example, if you have vectors \( \mathbf{a} \) and \( \mathbf{b} \), and their dot product \( \mathbf{a} \cdot \mathbf{b} = 0 \), they are orthogonal.
- This quality is crucial in areas like computer graphics and physics, where orthogonal vectors create independence among elements, making calculations more straightforward.
Vector Normalization
Vector normalization takes a vector and scales it to have a length (or magnitude) of one while maintaining its direction. The result is a unit vector. This is handy as it allows for simplifying vector operations, ensuring vectors are always treated equally regardless of their original size or scale.
To normalize a vector, you divide each component of the vector by its norm (or magnitude).
To normalize a vector, you divide each component of the vector by its norm (or magnitude).
- If you have vector \( \mathbf{v} = \begin{bmatrix} a & b & c \end{bmatrix} \), normalization makes it \( \mathbf{u} = \frac{\mathbf{v}}{||\mathbf{v}||} \).
- The squared length of the unit vector totals one, keeping the orientation of the original vector unaffected.
Dot Product
The dot product is a fundamental vector operation that provides crucial insights into the relationship between two vectors. It's essentially a way to multiply two vectors, which results in a scalar value.
This scalar can tell us whether two vectors are orthogonal.
This scalar can tell us whether two vectors are orthogonal.
- Calculated as: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).
- If the dot product is zero, the vectors are orthogonal.
- The dot product also informs us about the size of the angle between vectors and their alignment.
Vector Norm
The vector norm, often referred to as the vector's magnitude or length, is a measure of the "size" of the vector. It's calculated using the square root of the sum of the squares of its components.
The norm provides a straightforward measure of how long a vector is.
The norm provides a straightforward measure of how long a vector is.
- Calculated as: \( ||\mathbf{v}|| = \sqrt{a^2 + b^2 + c^2} \) for a vector \( \mathbf{v} = \begin{bmatrix} a & b & c \end{bmatrix} \).
- The norm is crucial when normalizing vectors and when verifying if a set of vectors are orthonormal.
- By comparing norms, you can easily adjust vectors to desirable lengths, such as making them unit-length and suitable for creating orthonormal sets.
