The equation relating a vector to a unit vector

[Mr Davis 97]

I am studying physics, and I see the equation $\hat{A} = \frac{\vec{A}}{A}$. What makes this relation obvious? It's quite obvious when one of the components of vector A is zero, but if both components are not zero, then what leads me to believe that this relation works every time?

 

[andrewkirk]

The equation should be written
$$\hat A=\frac{\vec A}{\left|\vec A\right|}$$
where $\left|\vec A\right|$ denotes the magnitude of vector $\vec A$ (sometimes written $\|\vec A\|$).
which is actually more easily understood as a scalar multiplication of a vector, viz:
$$\hat A=\left(\frac1{\left|\vec A\right|}\right)\ {\vec A}$$

It is true because it is the definition of the symbol $\hat A$.

If $\left|\vec A\right|=0$ then $\hat A$ has no meaning.

 

[Lucas SV]

Assuming you mean $A$ is the magnetude of $\overrightarrow{A}$, this is just a definition of the unit vector of $A$, for vectors of non-zero magnetude (otherwise the definition makes no sense). Are you asking why $\hat{A}$ is a unit vector? If you take the magnetude of $\frac{1}{A}\overrightarrow{A}$ you get $\frac{A}{A}=1$.

 

[Mr Davis 97]

If you take the magnetude of $\frac{1}{A}\overrightarrow{A}$ you get $\frac{A}{A}=1$.

Why?

 

[Lucas SV]

Why?

Suppose your vectors live in $n$ dimensions. Then the (Euclidean) norm, or magnetude of the vector $\overrightarrow{x}$ is defined as
$$
|(x_1,x_2,\cdots,x_n)|=\sqrt[]{x_1^2+x_2^2+\cdots+x_n^2}
$$
Then it is easy to show that $|k\overrightarrow{x}|=|k||\overrightarrow{x}|$, where $k$ is a real number (Do it!). Then just apply this result for the case $k=\frac{1}{|\overrightarrow{x}|}$.

 

[fresh_42]

Why?

1. Reread @andrewkirk's post.

2. $\hat{A}$ (for $\vec{A} \neq \vec{0}$) has always the same direction as $\vec{A}$, but its length is reduced to $1$.

3. Draw $\vec{A}$ with a ruler. Then $\hat{A}$ is the vector at the point $1$ of the ruler's scale. Now think of how you would express it in terms of $\vec{A}$.

 

[Mark44]

1. Reread @andrewkirk's post.

2. $\hat{A}$ (for $\vec{A} \neq \vec{0}$) has always the same direction as $\vec{A}$, but its length is reduced to $1$.

Or expanded to 1, if the magnitude of the vector is less than 1.

3. Draw $\vec{A}$ with a ruler. Then $\hat{A}$ is the vector at the point $1$ of the ruler's scale. Now think of how you would express it in terms of $\vec{A}$.