Let ℝ^3 have the standard dot product. Classify the...

Let $\mathbb{R}^3$ have the standard dot product. Classify the following sets of vectors as (i) basis, (ii) orthogonal basis, and/or (iii) orthonormal basis: (a) $\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right),\left(\begin{array}{r}1 \\ -1 \\ 1\end{array}\right),\left(\begin{array}{l}0 \\ 1 \\ 1\end{array}\right)$; (b) $\left(\begin{array}{c}-\frac{4}{13} \\ \frac{3}{5} \\ -\frac{48}{65}\end{array}\right),\left(\begin{array}{c}\frac{12}{13} \\ 0 \\ -\frac{5}{13}\end{array}\right),\left(\begin{array}{c}\frac{3}{13} \\ \frac{4}{5} \\ \frac{36}{65}\end{array}\right)$; (c) $\left(\begin{array}{r}0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{array}\right),\left(\begin{array}{r}-\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}\end{array}\right),\left(\begin{array}{r}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0\end{array}\right)$.

Let $\mathbb{R}^3$ have the standard dot product. Classify the following sets of vectors as
(i) basis, (ii) orthogonal basis, and/or (iii) orthonormal basis:
(a) $\left(\begin{array}{l}1 \\ 1 \\ 0\end{array}\right),\left(\begin{array}{r}1 \\ -1 \\ 1\end{array}\right),\left(\begin{array}{l}0 \\ 1 \\ 1\end{array}\right)$;
(b) $\left(\begin{array}{c}-\frac{4}{13} \\ \frac{3}{5} \\ -\frac{48}{65}\end{array}\right),\left(\begin{array}{c}\frac{12}{13} \\ 0 \\ -\frac{5}{13}\end{array}\right),\left(\begin{array}{c}\frac{3}{13} \\ \frac{4}{5} \\ \frac{36}{65}\end{array}\right)$;
(c) $\left(\begin{array}{r}0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{array}\right),\left(\begin{array}{r}-\frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}}\end{array}\right),\left(\begin{array}{r}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0\end{array}\right)$.

Key Concepts

Standard Dot Product

The standard dot product in ?³ is a bilinear form that takes two vectors and produces a scalar. This operation is fundamental to defining length (norms) and angles between vectors, and it is essential in determining properties like orthogonality and normalization.

Orthonormal Basis

An orthonormal basis is an orthogonal basis in which each vector is also of unit length (normalized to one). This normalization further simplifies computations, particularly in finding coordinates, projections, and in various numerical methods.

Basis

A basis in a vector space is a set of vectors that are both linearly independent and span the entire space. In ?³, any set of three linearly independent vectors forms a basis, ensuring that every vector in ?³ can be uniquely expressed as a linear combination of these basis vectors.

Orthogonal Basis

An orthogonal basis is a basis in which every pair of distinct vectors is orthogonal, meaning their dot product is zero. This property simplifies many calculations, such as expressing a vector in terms of the basis and computing coefficients in the linear combination.

Transcript

00:01 Hello there, so for this exercise we need to determine if the set of vectors that we have here is an orthogonal set or an orthomnormal set or neither okay, so for that just for addition we're going to call gamma omega sorry the set of uv and w or set of interest.

00:26 Okay, so we need to check if an omega in orthogonal set so for that we need to compute the inner product between all the the possible combinations between the vectors on the set that means uv and uv okay so that means uw and v we need to check that each of this inner product here are equals to zero that means that the set the elements of the set are pairwise or photo.

01:03 So in this case for u and v that corresponds to taking well the multiplication of the these canars that are multiplying to the vectors that means one over the square root of six time and then the multiplication of each component by each component of the vector the usual in your product in the clean space that means one minus one and and this is equal to zero so this in your product is equal to zero.

01:36 Then u with w.

01:38 So it happened the same here we got one half and then component by component you can see that here is minus one times one and here one times one that is equals to minus one plus one is also zero.

01:54 And the last one that we need to check is v with w.

01:59 So in this case this is equals to 1 over the square root of 6 times minus 1 minus 1 and here we have a problem because in this case this inner product is different from 0 because is equals to minus 2 over the square root of 6 which is different from 0 and therefore what can we say about this set omega is that omega is not orthogonal and therefore is not orthogonal.

02:38 That means that is nor orthogonal northonormal.

02:46 Okay.

02:48 The next one, the next set, is the set of these vectors.

02:56 So again we're going to define u as the set of you, to v and w.

03:03 Then we need to repeat the process that means calculating the inner product of you would be then you with w and finally v with w okay so this inner product is going to be equals to 1 over 9 times 4 plus 2 minus 2 which is clearly equals to 0...

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