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This vector is still orthogonal to the original vector $(x,y,z)$ as it was just scaled by a factor. It also has zero norm if and only if the norm of the original vector is 0. It also has zero norm if and only if the norm of the original vector is 0.
You would solve the 2x2 system to find a third orthogonal vector. Then for a fourth, you would fix two coordinates and have three unknowns, and now solve a 3x3 system. Similar logic shows that the fifth orthogonal vector would require solving a 4x4 system of equations.
An orthogonal basis can be used to decompose something into independent components. For example, the Fourier transform decomposes a time domain function into weights of sines and cosines. A triple in 3D space is a decomposition of a vector in 3D space along 3 orthogonal basis vectors. $\endgroup$ –
Two vectors are orthogonal if their inner product is zero. In other words $\langle u,v\rangle =0$. They are orthonormal if they are orthogonal, and additionally each vector has norm $1$. In other words $\langle u,v \rangle =0$ and $\langle u,u\rangle = \langle v,v\rangle =1$. Example. For vectors in $\mathbb{R}^3$ let
So if I would like to determine an orthogonal vector regarding: \begin{bmatrix}-1\\1\end{bmatrix} I just intuitively uses: $$\langle v,w\rangle=1 \cdot(-1)+1\cdot 1=0 $$ in order to arrive at \begin{bmatrix}1\\1\end{bmatrix} My problem is that I just dont know a mechanic way to solve for an orthogonal vector. It was more a educated guess.
Finding a basis and an orthogonal vector. 2. Finding all vectors orthogonal to a vector. 2.
Three non-colinear points are needed to define a plane, but you only really have two here. If you just want to find a couple of arbitrary vectors orthogonal vectors, just pick any third point, C, not colinear with A and B, then W = (A-C)×(B-C) (where × is the cross product) is orthogonal to the line segment connecting A and B since any plane containing A and B also contains the line segment ...
Also, orthogonal set and linearly independent set both generate the same subspace. (Is that right?) Then orthogonal $\rightarrow$ linearly independent but orthogonal $\nleftarrow$ linearly independent is that right? One more question. For T/F, Every orthogonal set is linearly independent (F) Every orthonormal set is linearly independent (T) Why?
I'm currently in a linear algebra course and I am just looking for feedback or some sort of verification of what I'm doing is correct.
Here, the result follows from the definition of "mutually orthogonal". A set of vectors is said to be mutually orthogonal if the dot product of any pair of distinct vectors in the set is 0. This is the case for the set in your question, hence the result.