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}##.