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Chapter 4: Problem 2

Determine whether the following set of vectors is orthogonal. If it is orthogonal, determine whether it is also orthonormal. $$ \left[\begin{array}{r} 1 \\ 2 \\ -1 \end{array}\right],\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{r} -1 \\ 1 \\ 1 \end{array}\right] $$ If the set of vectors is orthogonal but not orthonormal, give an orthonormal set of vectors which has the same span.

Short Answer

Expert verified
The set is orthogonal but not orthonormal. The orthonormal set is \(\left[ \begin{array}{r} \frac{1}{\sqrt{6}} \ \frac{2}{\sqrt{6}} \ \frac{-1}{\sqrt{6}} \end{array}\right], \left[ \begin{array}{r} \frac{1}{\sqrt{2}} \ 0 \ \frac{1}{\sqrt{2}} \end{array}\right], \left[ \begin{array}{r} \frac{-1}{\sqrt{3}} \ \frac{1}{\sqrt{3}} \ \frac{1}{\sqrt{3}} \end{array}\right] \).

Step by step solution

01

Define Orthogonality

Two vectors are orthogonal if their dot product is zero. To determine if the given set of vectors is orthogonal, each pair must have a dot product equal to zero.
02

Compute Dot Products

Calculate the dot product of each pair of vectors:\(\left[ \begin{array}{r} 1 \ 2 \ -1 \end{array}\right] \cdot \left[ \begin{array}{r} 1 \ 0 \ 1 \end{array}\right] = (1 \cdot 1) + (2 \cdot 0) + (-1 \cdot 1) = 1 - 1 = 0\)\(\left[ \begin{array}{r} 1 \ 2 \ -1 \end{array}\right] \cdot \left[ \begin{array}{r} -1 \ 1 \ 1 \end{array}\right] = (1 \cdot -1) + (2 \cdot 1) + (-1 \cdot 1) = -1 + 2 - 1 = 0\)\(\left[ \begin{array}{r} 1 \ 0 \ 1 \end{array}\right] \cdot \left[ \ begin{array}{r} -1 \ 1 \ 1 \end{array}\right] = (1 \cdot -1) + (0 \cdot 1) + (1 \cdot 1) = -1 + 0 + 1 = 0\)
03

Determine Orthogonality

Since all the dot products are zero, the set of vectors is orthogonal.
04

Check for Orthonormality

A set of vectors is orthonormal if it is orthogonal and each vector has a magnitude of 1. Calculate the magnitude of each vector:\(\left[ \begin{array}{r} 1 \ 2 \ -1 \end{array}\right] \) magnitude: \( \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6} \)\(\left[ \begin{array}{r} 1 \ 0 \ 1 \end{array}\right] \) magnitude: \( \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2} \)\(\left[ \begin{array}{r} -1 \ 1 \ 1 \end{array}\right] \) magnitude: \( \sqrt{(-1)^2 + 1^2 + 1^2} = \sqrt{3} \)
05

Normalize the Vectors

Since the vectors do not have a magnitude of 1, normalize each vector:\(\left[ \begin{array}{r} 1 \ 2 \ -1 \end{array}\right] \) normalized: \( \frac{1}{\sqrt{6}} \left[ \begin{array}{r} 1 \ 2 \ -1 \end{array}\right] = \left[ \begin{array}{r} \frac{1}{\sqrt{6}} \ \frac{2}{\sqrt{6}} \ \frac{-1}{\sqrt{6}} \end{array}\right] \)\(\left[ \ begin{array}{r} 1 \ 0 \ 1 \end{array}\right] \) normalized: \( \frac{1}{\sqrt{2}} \left[ \begin{array}{r} 1 \ 0 \ 1 \end{array}\right] = \left[ \begin{array}{r} \frac{1}{\sqrt{2}} \ 0 \ \frac{1}{\sqrt{2}} \end{array}\right] \)\(\left[ \begin{array}{r} -1 \ 1 \ 1 \end{array}\right] \) normalized: \( \frac{1}{\sqrt{3}} \left[ \ begin{array}{r} -1 \ 1 \ 1 \end{array}\right] = \left[ \begin{array}{r} \frac{-1}{\sqrt{3}} \ \frac{1}{\sqrt{3}} \ \frac{1}{\sqrt{3}} \end{array}\right] \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dot Product
The dot product is a fundamental concept in vector mathematics. It is calculated by multiplying corresponding components of two vectors and then summing these products.
For vectors \(\textbf{a} = [a_1, a_2, a_3]\) and \(\textbf{b} = [b_1, b_2, b_3]\), the dot product is given by:
\( \textbf{a} \bullet \textbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)
It can be interpreted as a measure of how much one vector extends in the direction of another.
If the dot product is zero, the vectors are orthogonal (perpendicular).
Orthonormal Set
An orthonormal set is a collection of vectors that are both orthogonal and normalized. Orthogonal means that each pair of vectors in the set has a dot product of zero. Normalized means that each vector in the set has a magnitude (or length) of one.
This property is significant because orthonormal sets are easier to work with in various mathematical computations and applications.
If you have a set of orthogonal vectors, you can convert it into an orthonormal set by normalizing each vector.
This helps in simplifying problems and is used extensively in fields like computer graphics and data science.
Vector Normalization
Normalization is the process of converting a vector into a unit vector. A unit vector has a magnitude (length) of one.
To normalize a vector \(\textbf{v} = [v_1, v_2, v_3]\), you divide each component by the magnitude of the vector. The formula is:
\( \text{Normalized} \textbf{v} = \frac{1}{||\textbf{v}||} \textbf{v} = \frac{1}{\text{magnitude of } \textbf{v}} [v_1, v_2, v_3] \)
For example, if a vector \(\textbf{v}\) has a magnitude of \(\textbf{\rho}\), the normalized vector would be:
\( \textbf{v}_{\text{normalized}} = [ \frac{v_1}{\textbf{\rho}}, \frac{v_2}{\textbf{\rho}}, \frac{v_3}{\textbf{\rho}} ] \)
Normalize each vector to form an orthonormal set.
Vector Magnitude
The magnitude of a vector is essentially its length. For a vector \(\textbf{v} = [v_1, v_2, v_3]\), the magnitude is calculated using the Pythagorean theorem:
\[ ||\textbf{v}|| = \text{sqrt}(v_1^2 + v_2^2 + v_3^2) \]
Understanding a vector鈥檚 magnitude is crucial for normalization and for determining if sets of vectors can be converted into orthonormal sets.
It helps in comparing the sizes of vectors and is pivotal in various applications like physics, computer graphics, and machine learning. Remember, for vectors to be part of an orthonormal set, their magnitude must be 1.

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Most popular questions from this chapter

Here are some vectors in \(\mathbb{R}^{4}\). $$ \left[\begin{array}{r} 1 \\ 3 \\ -1 \\ 1 \end{array}\right],\left[\begin{array}{r} -\frac{3}{2} \\ -\frac{9}{2} \\ \frac{3}{2} \\ -\frac{3}{2} \end{array}\right],\left[\begin{array}{r} 1 \\ 4 \\ -1 \\ 1 \end{array}\right],\left[\begin{array}{r} 2 \\ -1 \\ -2 \\ 2 \end{array}\right],\left[\begin{array}{c} 1 \\ 4 \\ 0 \\ 1 \end{array}\right] $$ Thse vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Here are some vectors in \(\mathbb{R}^{4}\). $$ \left[\begin{array}{r} 1 \\ 1 \\ -1 \\ 1 \end{array}\right],\left[\begin{array}{r} 1 \\ 2 \\ -1 \\ 1 \end{array}\right],\left[\begin{array}{r} 1 \\ -2 \\ -1 \\ 1 \end{array}\right],\left[\begin{array}{l} 1 \\ 2 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{r} 1 \\ -1 \\ -1 \\ 1 \end{array}\right] $$ Thse vectors can't possibly be linearly independent. Tell why. Next obtain a linearly independent subset of these vectors which has the same span as these vectors. In other words, find a basis for the span of these vectors.

Find the rank of the following matrix. Also find a basis for the row and column spaces. $$ \left[\begin{array}{rrrrrr} 1 & 0 & 3 & 0 & 7 & 0 \\ 3 & 1 & 10 & 0 & 23 & 0 \\ 1 & 1 & 4 & 1 & 7 & 0 \\ 1 & -1 & 2 & -2 & 9 & 1 \end{array}\right] $$

City A is located at the origin (0,0) while city \(B\) is located at (300,500) where distances are in miles. An airplane flies at 250 miles per hour in still air. This airplane wants to fly from city \(A\) to city \(B\) but the wind is blowing in the direction of the positive \(y\) axis at a speed of 50 miles per hour. Find a unit vector such that if the plane heads in this direction, it will end up at city \(B\) having flown the shortest possible distance. How long will it take to get there?

The wind blows from the West at 30 kilometers per hour and an airplane which flies at 400 kilometers per hour in still air is heading North East. Find the velocity of the airplane and its position after two hours.
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