Question

We often write vectors in \(\mathbb{R}^{n}\) as row \(\mathrm{n}\) -tuples. In each case, show that the set of vectors is orthogonal in \(\mathbb{R}^{4}\). a. \(\\{(1,-1,2,5),(4,1,1,-1),(-7,28,5,5)\\}\) b. \(\\{(2,-1,4,5),(0,-1,1,-1),(0,3,2,-1)\\}\)

Short answer

Both sets of vectors are orthogonal in $ }}{ R}^{4} .

Step-by-step solution

Define the concept of orthogonal vectors

Two vectors in $ { R}^{4} $ are orthogonal if their dot product is equal to zero. The dot product of two vectors $ {a}=(a_1, a_2, a_3, a_4) $ and $ {b}=(b_1, b_2, b_3, b_4) $ is given by $ a ullet b a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 a_1b_1 + a_2b_2 + a_3b_3 + a_4b_4 $.

Calculate dot products for set (a)

For set (a), let's calculate the dot products of each pair: 1. def a:b = (1, −1, 2, 5) a_second b = (4,1,1,−1) a −1, a_second 2, a_second 5) a cdef a b = 1×4 + (-1)×1 + 2×1 + 5×(-1) = 4 - 1 + 2 - 5 = 0 a c_2:def d_for a, def bc = (b) 2. def def For For for (1, −1, b): def a 2, 5), using a for a: b = −1, (a 5agsrf (-7, 28, 5, 5) b = 1×(-7) + (-1)×28 + 2×5 + 5×5 = -7 - 28 + 10 + 25 = 0 3. a= (4,a ,1, (a:-(−7, 28, 1) 5), a 1) b a_for {4) c = (4(4: =b)= 4×( = −7) + (1) def × 28) + (1)×5 + again FOR b= -28 + 28 + 5 - 5 = this it's - 28 + c c for a = Again: 0 Therefore, all pairs of vectors in set (a) are orthogonal.

Calculate dot products for set (b)

For set (b), calculate the dot products: 1. a For= (2) b a=(2,−1,4,5) a, ,0) =(a dot a b (0,−1,1 −1) a a dot p = o (b)= 2×0 + (-1)×(-1) + 4×1 + 5×(-1) = 0 + 1 + 4 - 5 = 0 2. and (2,−1,4,5) a:(a, a , a)(a (0,3,2 ⟂ ⟂cot p(b)=-1) a a: b) b) = Dot = 2×0 + (-1)×3 + 4×2 + 5×(-1) = 0 - 3 + 8 - 5 = 0 3. a (b: a b) (0,−1,(0 -d, 1)(1 ⟂ 0⟂ multipl b:-(a i.: ⟂(b) b =(0, , po)0 ⟂t busing a (0,)which and proves (3,i 3, 2) dot b=, −1)b 0×0 +(0)×3 + 1×2 + (-1)×(-1) = a 0 + 0+2+1 :b 0a=using a Also, 0 ) 2: Therefore, all pairs of vectors in set (b) are orthogonal as well.

Key concepts

Dot Product

The dot product is a fundamental operation in vector algebra. It measures how much one vector goes in the direction of another.
Understanding the dot product is crucial for determining if two vectors are orthogonal. Two vectors are orthogonal when their dot product is zero.

• Imagine vectors \(\mathbf{a}\) and \(\mathbf{b}\). If \(\mathbf{a} \cdot \mathbf{b} = 0\), then they are orthogonal.
• In mathematical terms, the dot product \(\mathbf{a} \cdot \mathbf{b}\) is calculated as \(a_1b_1 + a_2b_2 + \ldots + a_nb_n\) where \(a_i\) and \(b_i\) are components of the vectors.
• This operation involves multiplying corresponding components and summing the results.

Orthogonal vectors have significant applications in engineering and physics as they represent independent directions. Recognizing when vectors are orthogonal helps in simplifying complex equations and systems.

Vectors in \(\mathbb{R}^{n}\)

Vectors in \(\mathbb{R}^{n}\) are essential in mathematics for representing multi-dimensional quantities. Each vector has components which suggest its position in an \(n\)-dimensional space.
Working with these vectors means identifying their length (magnitude), direction, and dimension.

• For instance, a vector in \(\mathbb{R}^{4}\) is written as \((a_1, a_2, a_3, a_4)\).
• This notation means the vector has four components, each representing a coordinate in 4-dimensional space.

Understanding vectors in \(\mathbb{R}^{n}\) is important for several reasons:

• They can represent anything from velocity and force in physics to data points in machine learning.
• Handling different dimensions is key to solving complex real-world problems.

Moreover, recognizing the orthogonality among such vectors can greatly enhance problem-solving efficiency.

Vector Algebra

Vector algebra involves operations like addition, subtraction, and scalar multiplication, as well as the dot and cross product.
Vector algebra enables us to perform calculations that are common in physics and engineering, such as force decomposition and projections.

• Addition of vectors \(\mathbf{a} + \mathbf{b}\) results in a new vector by adding corresponding components.
• Scalar multiplication \(c \cdot \mathbf{a}\) simply scales every component of \(\mathbf{a}\) by \(c\).
• Subtraction finds the difference, much like how addition works but in reverse.
• The dot product, as discussed, helps in finding orthogonality and projections onto other vectors.

Learning vector algebra is fundamental for exploring more advanced topics in mathematics and its applications.

• It provides the basics for geometry studies, calculus in vector fields, and even in optimizing functions in algorithms.
• Understanding it lays the groundwork for a deeper comprehension of how multi-dimensional analysis works in various scientific fields.

Therefore, mastering vector algebra opens doors to exploring deeper mathematical concepts and applications.