Show $a·b = |a| × |b| \cos(\theta)$ geometrically and by using no algebraic arguments at all

I want to know if there is a more natural way of deriving $a·b = |a| × |b| \cos(\theta)$ without using algebraic identities and looking at a figure instead. I am familiar with the algebraic method.

-
How do you define $a\cdot b$ geometrically? –  Nameless Dec 30 '12 at 8:25

The dot product of two Euclidean vectors $a$ and $b$ is defined by the same formula you noted. Also, the scalar projection of a Euclidean vector $a$ onto a Euclidean vector $b$ is known by

$$a\cdot b=|a|\cos\theta$$

where $θ$ is the angle between two vectors $a$ and $b$.For having a geometric definition of the dot product, this can be rewritten as:

$$a_b=|a|\cdot\widehat{\mathbf b}$$

where $\widehat{\mathbf b} = b/| b|$ is the unit vector in the direction of B. So The dot product is thus characterized geometrically by

$$a\cdot b = a_b|b|$$.

-
Nice picture (and answer!) Did you make the picture? +1 –  amWhy Feb 27 '13 at 0:29
@amWhy: I think I copied it from somewhere in web in that time. –  Babak S. Feb 27 '13 at 2:37
That's fine...I just know you do a lot of graphics "home-made"! :-) –  amWhy Feb 27 '13 at 2:38

For a unit vector $\hat{\bf w}$, the dot product $\bf{v}\cdot \hat{\bf{w}}$ is the length of the orthogonal projection of $\bf v$ onto $\hat{\bf w}$. By drawing a right triangle you can see that the length of this projection is given by $|\textbf{v}|\cos\theta$, where $\theta$ is the angle between the vectors.

Notice that the dot product is linear in the first argument. If you accept as given that the dot product should be symmetric, we now get the full formula for arbitrary $\textbf{w}$: \begin{align*} \textbf{v}\cdot\textbf{w} &= |\textbf{v}|(\hat{\textbf{v}}\cdot \textbf{w})\\ &= |\textbf{v}|(\textbf{w}\cdot\hat{\textbf{v}})\\ &= |\textbf{v}||\textbf{w}|\cos\theta. \end{align*}

-