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What exactly is the difference between $\vec{a} $ and $\lVert\vec{a}\rVert$ ?

I know $\lVert\vec{a}\rVert$ is the magnitude of a vector $\vec{a} $ , but what is $\vec{a} $ then exactly? I know it's a vector, but how can you represent it? Let's say $\vec{a} $ is a vector in $\mathbb{R}^2$, if you draw $\vec{a} $ in $\mathbb{R}^2$, don't you automatically draw $\lVert\vec{a}\rVert$ ?

Also, what is the difference between $ \lVert\vec{a}\rVert \cdot \big\lVert\vec{b}\big\rVert $ and $\vec{a} \cdot \vec{b} $? To me it seems like multiplying the same, but I know it isn't.

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$\vec{a}$ has magnitude and direction thus a real number isn't enough to represent it. Its magnitude on the other end is a real number. The rest should follow easily if you are reading a book. – nt.bas Oct 7 '12 at 13:36
Try thinking of $\vec{a}$ as a point $(x,y)$ in $\mathbb{R}^2$ and $||\vec{a}||$ as the distance of that point from the origin, i.e. $\sqrt{x^2+y^2}$. – Shahab Oct 7 '12 at 13:45
up vote 5 down vote accepted

$\vec{a}$ is a vector, i.e. some element $(a_1,a_2)$ of $\mathbb{R}^2$ in your case. You draw it by drawing an arrow from the point $(0,0)$ to the point $(a_1,a_2)$.

$||\vec{a}||$ is the length of the vector $\vec{a}$, i.e. the length of the arrow from $(0,0)$ to $(a_1,a_2)$. $||\vec{a}||$ is thus a positive real number, i.e. it is an element of $\mathbb{R}$ (and is $\geq 0$), not of $\mathbb{R}^2$.

$||\vec{a}||\cdot||\vec{b}||$ is the product of the lengths of $\vec{a}$ and $\vec{b}$. It is thus also a positive real number. For a vector $\vec{a} = (a_1,a_2)$ the length $||\vec{a}||$ is $\sqrt{a_1^2 + a_2^2}$.

$\vec{a}\vec{b}$, sometimes also written $\left<\vec{a},\vec{b}\right>$ is the scalar product of $\vec{a}$ and $\vec{b}$. It's also an element of $\mathbb{R}$, not of $\mathbb{R}^2$. For two vectors $\vec{a} = (a_1,a_2)$ and $\vec{b} = (b_1,b_2)$ the definition of the scalar product is $\vec{a}\vec{b} = a_1b_1 + a_2b_2$.

Note that there is a connection between the two. By the definition of the scalar product above, $\vec{a}\vec{a}$ is $a_1^2 + a_2^2$. This is the same as $||\vec{a}||^2$.

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Amazing answer. Clear and concise! – JohnPhteven Oct 7 '12 at 13:54
Concise? This is the longest answer, so far... Not that length has anything to do with quality, mind you. – Did Oct 7 '12 at 14:09

Basically a vector has a length and a direction. When you write $\vec{a}$ you mean the whole vector. Whereas $||\vec{a}||$ denotes only the length of the vector.

For example $\vec{a}=(2,1) \in \mathbb{R}^2$. The length of this vector is $||\vec{a}||=\sqrt{2^2+1^2}=\sqrt{5}$.

When you look at $\vec{b}=(-2,1)$ we still have the same length ($||\vec{b}||=\sqrt{(-2)^2+1^2}=\sqrt{5}$) but the direction is different.

For $\vec{a} . \vec{b} $ or better: $\vec{a} \cdot \vec{b} $ you have to be careful. This is a special new operation, called scalar product or dot product. It has not much to do with the usual multiplication:

$$(2,1) \cdot (-2,1) = 2*(-2)+1*1=-3 \neq 5 =||(2,1)||*||(-2,1)||.$$

In some way the scalar product measures how the two vectors lie to each other in the plane (or in $\mathbb{R}^n$).

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If $\vec a=(1,0)$, then $\|\vec a\|=1$. One sees that $(1,0)\ne1$, that is, $\vec a\ne\|\vec a\|$. In fact, the objects $\vec a$ and $\|\vec a\|$ do not even belong to the same set since the first one belongs to $\mathbb R^2$ and the second one to $\mathbb R$.

If $\vec b=(0,1)$, then $\vec a\cdot\vec b=0$ while $\|\vec a\|\cdot\|\vec b\|=1$ hence $\vec a\cdot\vec b\ne\|\vec a\|\cdot\|\vec b\|$. Now, $\vec a\cdot\vec b$ and $\|\vec a\|\cdot\|\vec b\|$ are both real numbers but they can differ (and they usually do).

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Geometrically, $\vec{a}$ is a direction with length, $||\vec{a}||$ is only length and $\vec{a} \cdot \vec{b} / ||\vec{b}||$ is the length of projection of $\vec{a}$ onto the line given by $\vec{b}$.

Algebraically, $\vec{a} \in \mathbb{R}^n$ is a row of $n$ numbers, $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + ... +a_nb_n = ||\vec{a}||\cdot||\vec{b}||\cdot \sin \alpha$, where $\alpha$ is the angle between the vectors and $||\vec{a}|| = \vec{a} \cdot \vec{a}$.

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