How do Unit Vectors Behave in Triangular Coordinate Systems?

[cxcxcx0505]

How does it look like if three unit vectors at a point such that each vector is tangent to one of the axes in 3-Dimensional Cartesian System? Can anyone illustrate about this? Thanks.

 

[phinds]

How does it look like if three unit vectors at a point such that each vector is tangent to one of the axes in 3-Dimensional Cartesian System? Can anyone illustrate about this? Thanks.

It looks exactly like the standard drawing of a 3D graph basis, just with the center offset accordingly (unless your point is AT the origin, in which case it is exactly the standard drawing)

 

[cxcxcx0505]

It looks exactly like the standard drawing of a 3D graph basis, just with the center offset accordingly (unless your point is AT the origin, in which case it is exactly the standard drawing)

Does it mean that the vector is parallel to the axis?

 

[Mark44]

Does it mean that the vector is parallel to the axis?

Which vector and which axis?

You mentioned three unit vectors and three axes in your OP.

 

[cxcxcx0505]

Which vector and which axis?

You mentioned three unit vectors and three axes in your OP.

each unit vector parallel to each axes? if I have three unit vector u, v ,w, and axes x , y, z , then u parallel to x , v parallel to y , w parallel to z?

 

[Mark44]

If each one is tangent to an axis, then it is parallel to that axis.

 

[cxcxcx0505]

I'm reading "An Introduction to Tensors for Students of Physics and Engineering"

Now, some imagination is required. Let’s return to the 3-dimensional Cartesian system. At any
point P, we can specify three local axes and three local planes determined by these axes. In
accordance with strict definitions, the axes must be mutually perpendicular and, by extension, so
must the planes. Now, choose three unit vectors at P such that each vector is tangent to one of
the axes. Such a triple is usually designated (i, j, k). Any vector V at P can then be written
V = αi + βj + γk
where α, β, and γ are the usual x, y, and z scalar components of the vector.
Now suppose that we had chosen unit vectors perpendicular to each of the planes rather than
tangent to each of the coordinate axes. Let’s do so and call the resulting triple (i*, j*, k*). Again,
any vector V at P can be written
V = α*i* + β*j* + γ*k*
where α*, β*, and γ* are the scalar components of the vector referred to the i*, j*, k* triple.
There is nothing surprising in what we have just done, and our representation is satisfactory
provided we ensure that
αi + βj + γk = α*i* + β*j* + γ*k*.
But, you might argue that what we have done is trivial since it is apparent from geometry that the
two unit vector triples comprise the same set; i.e., that
i = i*
j = j*
k = k*.
Still, we used two distinct approaches to defining a unit vector triple at P. Should we expect
these approaches to produce so tidy a result in all cases? The answer is very definitely “NO”!
To understand why the answer is “NO,” let’s modify our Cartesian system so that the axes are no
longer mutually orthogonal – for example, so that they meet at 60°. In this case, the origin lies at
a vertex of a tetrahedron, and the axes lie along three of the edges. (Such coordinate systems are actually used in engineering and crystallography and are called triangular coordinate systems.) It should be intuitive that (i, j, k) and (i*, j*, k*) are now two different sets of unit vectors.
Specifically, i and i* now meet at an angle of 60°, as do j and j*, and k and k*. Thus, while they
are all unit vectors, they specify different sets of directions, and the choice of which set to use in
a given calculation must be a matter of expediency.

I cannot understand in the triangular coordinate systems, how i and i* now meet at an angle of 60°?