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In the notation:

Vector(v) = ai + bj + ck

Aren't a, b, c the components of vector? I looked on Wikipedia which says they "are the magnitudes of the components of v."

Does it mean the same?

If not then how can I convert the components of vector given in set notation:

$v = (c_x, c_y, c_z)$.

Where, $c_x, c_y, c_z$ are the components into the $i, j, k$ notation?

user963241
  • 739
  • After a small amount of Googling I've seen that some people refer to $a$, $b$, and $c$ as the components of $\mathbf v$ (wrt the standard basis) and some people refer to $a\mathbf i$, $b\mathbf j$, and $c\mathbf k$ as the components. Wikipedia obviously takes the second meaning and thus the magnitude of each of those components is $|a|$, $|b|$, and $|c|$ respectively. –  Nov 07 '15 at 17:57
  • For Wikipedia (and many people), the componets are the vectors $a \mathbf i$, $b\mathbf j$ and $c\mathbf k$. $a,b,c$ are its coordinates. – Bernard Nov 07 '15 at 17:57
  • If I have components given: $c_x, c_y, c_z$ how can I convert it into $i, j, k$ notation? – user963241 Nov 07 '15 at 18:01
  • In $\Bbb R^3$ the standard basis ${\mathbf i, \mathbf j, \mathbf k}$ is defined by $\mathbf i=(1,0,0)$, $\mathbf j = (0,1,0)$, and $\mathbf k=(0,0,1)$. Thus $v=(c_x, c_y,c_z) = c_x\mathbf i + c_y\mathbf j + c_z\mathbf k$. –  Nov 07 '15 at 18:04
  • So, if I have a vector $AB$, $A = (2,3)$ and $B (7, 6)$ then components will be $(7-2, 6-3) = (5, 3)$ then I can simply write it in $i, j, k$ notation: $5i + 3j + 0k$ right? – user963241 Nov 07 '15 at 18:25
  • $\mathbf k$ doesn't exist in $\Bbb R^2$ so you'd just write it as $5\mathbf i + 3\mathbf j$. On the other hand, if $A$ and $B$ are the points $A(2,3,0)$ and $B(7,6,0)$, then $(\vec {AB})_{\mathcal E}$ does equal $5\mathbf i + 3\mathbf j + 0\mathbf k$. Though even then the $0\mathbf k$ is usually left off because it's just the zero vector. –  Nov 07 '15 at 18:37
  • For more information about vectors, I've (as well as many, many others) written several answers here that might help you understand these concepts a little better. Two that I'll single out are this one and this one. –  Nov 07 '15 at 18:43

1 Answers1

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Consider a vector $\mathbf v \in V$ where $V$ is a $3$-dimensional real vector space. There is a theorem which tells us that every vector space has at least one basis. That allows us to say there exists a basis $S=\{\mathbf s_1, \mathbf s_2, \mathbf s_3\}$ of $V$. By the definition of a basis, we can write $\mathbf v$ as a linear combination of $S$:

$$\mathbf v = v_1\mathbf s_1 + v_2\mathbf s_2 + v_3\mathbf s_3$$

After choosing a basis for $V$ we can make use of another very useful theorem in linear algebra: every $n$-dimensional real vector space is isomorphic to $\Bbb R^n$, where $\Bbb R^n$ is the space of all ordered "tuples" of real numbers: for instance $\Bbb R^3 = \{(a,b,c) \mid a,b,c \in \Bbb R\}$. What that means for us is that we can freely convert any vectors in our abstract real vector space $V$ to vectors in $\Bbb R^3$ and vice versa. The "natural" way to do so is $$\mathbf v = v_1\mathbf s_1 + v_2\mathbf s_2 + v_3\mathbf s_3 \leftrightarrow (v_1, v_2, v_3) = (\mathbf v)_S$$

The fact that $V$ and $\Bbb R^3$ are isomorphic as vector spaces means that if we'd like to perform a linear transformation $T$ of a vector $\mathbf v\in V$ we can instead first convert $\mathbf v \to (\mathbf v)_S$, perform the transformation in $\Bbb R^3$, and then convert back $T((\mathbf v)_S) \to T(\mathbf v)$. Either way we'll get the same answer.

Here's an example: let $\mathbf v= 2\mathbf s_1 + 3\mathbf s_3$ and $\mathbf w=\mathbf s_1 -2\mathbf s_2 + \mathbf s_3$. Consider the linear transformation $T(\mathbf x) = 2\mathbf x + \mathbf w$. Then $$\begin{align}T(\mathbf v) &= 2\mathbf v + \mathbf w \\ &= 2(2\mathbf s_1 + 3\mathbf s_3) + (\mathbf s_1 -2\mathbf s_2 + \mathbf s_3) \\ &= (4+1)\mathbf s_1 +(0-2)\mathbf s_2 + (6+1)\mathbf s_3 \\ &= 5\mathbf s_1 -2\mathbf s_2 + 7\mathbf s_3\end{align}$$ vs $$\begin{align}T(\mathbf v) \rightarrow T([\mathbf v]_S) &= 2(\mathbf v)_S + (\mathbf w)_S \\ &= 2(2,0,3) + (1,-2,1) \\ &= (4+1,0-2,6+1) \\ &= (5,-2,7) \\ &\rightarrow 5\mathbf s_1 -2\mathbf s_2 + 7\mathbf s_3\end{align}$$

Note: Technically $T: V\to V$ and $T:\Bbb R^3\to \Bbb R^3$ are different transformations -- and thus should probably have different names -- but usually we're not that observant.

As for the nomenclature, apparently some people use the term "components of a vector $\mathbf v$ (with respect to $S$)" to mean the scalars $v_1$, $v_2$, and $v_3$ whereas others call the components $v_1\mathbf s_1$, $v_2\mathbf s_2$, and $v_3\mathbf s_3$ -- distinguishing $v_1$, $v_2$, and $v_3$ as the coordinates of $\mathbf v$.

And for the notation, for historical reasons (look up the quaternions for more info) the "standard" basis of $\Bbb R^3$ is often labeled $\mathcal E = \{\mathbf i, \mathbf j, \mathbf k\}$ where $\mathbf i=(1,0,0)$, $\mathbf j = (0,1,0)$, and $\mathbf k= (0,0,1)$ by definition.

Thus $$(\mathbf v)_S = (v_1, v_2, v_3) = v_1\mathbf i + v_2\mathbf j + v_3\mathbf k$$