In each part, determine whether the set of vectors is...

In each part, determine whether the set of vectors is orthogonal and whether it is orthonormal with respect to the Euclidean inner product on $R^{3}$ (a) $\left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right),\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$ (b) $\left(\frac{2}{3},-\frac{2}{3}, \frac{1}{3}\right),\left(\frac{2}{3}, \frac{1}{3},-\frac{2}{3}\right),\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$ (c) $(1,0,0),\left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),(0,0,1)$ (d) $\left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right),\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, 0\right)$

In each part, determine whether the set of vectors is orthogonal and whether it is orthonormal with respect to the Euclidean inner product on $R^{3}$
(a) $\left(\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right),\left(-\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}\right)$
(b) $\left(\frac{2}{3},-\frac{2}{3}, \frac{1}{3}\right),\left(\frac{2}{3}, \frac{1}{3},-\frac{2}{3}\right),\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$
(c) $(1,0,0),\left(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),(0,0,1)$
(d) $\left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right),\left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}, 0\right)$

Key Concepts

Orthonormality

A set of vectors is orthonormal if it is both orthogonal and every vector in the set is normalized. This means that not only are the vectors mutually perpendicular (with zero inner product between distinct vectors), but each vector also has a unit norm, which is a highly desirable property for bases in vector spaces.

Normalization

Normalization is the process of scaling a vector so that its norm (or length) becomes 1. A normalized vector is often referred to as a unit vector, and this property is crucial when considering orthonormal sets where every vector must have a length of 1.

Orthogonality

A set of vectors is called orthogonal if every pair of distinct vectors within the set has an inner product of zero. This property is fundamental in simplifying many problems in linear algebra, such as projections and decompositions, because orthogonal vectors do not interfere with each other’s directions.

Euclidean Inner Product

This is the operation used to calculate the dot product between two vectors, defined for two vectors x and y in R^n as the sum of the products of their corresponding components. It provides a measure of how 'aligned' two vectors are and is key for determining orthogonality and lengths (norms) of vectors.

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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