What exactly is the number j ? I am not talking about the j that electrical engineers use. I am talking about the J which is used for describing three-dimensional numbers.

Also, why is it not possible to compute i*j? Why can’t we do computations with three-dimensional numbers?

For your first question, $\mathbf{j}$ is typically notation for the vector $\langle 0, 1, 0\rangle$. There are two others (with $1$s in the other positions) which give a basis for $3$-dimensional Euclidean space.

On your second question: to be transparent, you can in fact multiply three-component vectors. You would do it as $$ \langle a, b, c \rangle \langle d, e, f \rangle = \langle ad, be, cf \rangle. $$ This is known as the direct product ring structure on $\mathbb{R}^3$. This has most of the nice properties of algebraic settings with a $+$ and a $\times$, but not all. You need a grounding in abstract algebra to really get this. But as examples, this operation would satisfy friendly-looking properties like $$ \mathbf{u} \mathbf{v} = \mathbf{v} \mathbf{u} $$ and $$ \mathbf{u}(\mathbf{v}+\mathbf{w}) = \mathbf{u}\mathbf{v} + \mathbf{u}\mathbf{w} $$ and so on.

However, the real point is that this sort of multiplication simply isn't useful. So, can you do it? Yes. Is it useful? No. This is why you don't commonly see it.

October 10, 2019 01:32 AM

In three dimensional space, there are three standard unit vectors which are mutually perpendicular.

$$i=(1,0,0),j=(0,1,0),k=(0,0,1)$$

Thus they are the unit vectors along the coordinate axes.

You may find the dot product or the cross product of them with each other or with other vectors in your space.

As you know we also have $$ (a,b,c)= ai+bj+ck$$ for any vector $(a,b,c)$

October 10, 2019 01:38 AM

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