Dot product of two vectors on plane could be defined as product of lengths of those vectors and cosine of angle between them.

In cartesian coordinates dot product of vectors with coordinates $(x_1, y_1)$ and $(x_2, y_2)$ is equal to $x_1x_2 + y_1y_2$.

How to prove it?

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    $\begingroup$ How do you define the length of a vector without dot product? |v| = √(v∙v)... without this definition your first definition doesn't make sense. You can, however, think simply in terms of cartesian coordinates and prove that x1*x2 + y1*y2 = (x1^2 + y1^2)*(x2^2 + y2^2) * cos (theta). $\endgroup$ – Sklivvz Aug 1 '10 at 13:25

I suppose you want to prove that two your definitions of dot product are the same. We start with definition of dot product as $(\vec{u}, \vec{v}) = |\vec{u}| |\vec{v}| \cos \theta$. We start with definition of dot product as $(\vec{u}, \vec{v}) = |\vec{u}| |\vec{v}| \cos \theta$ and prove that it also satisfies $(\vec{u}, \vec{v}) = x_1 x_2 + y_1 y_2$.

At first you can prove that dot product is linear: $(\vec{v_1}, \vec{v_2} + \alpha \vec{v_3}) = (\vec{v_1}, \vec{v_2}) + \alpha (\vec{v_1}, \vec{v_3})$. This is true because $(\vec{v_1}, \vec{v_2})$ is equal to the product of $|\vec{v_1}|$ and projection of $\vec{v_2}$ on $\vec{v_1}$. Projection of sum of vectors is equal to sum of projections. Hence dot product is linear.

Let $\vec{e_1}$ and $\vec{e_2}$ be vectors with coordinates $(1, 0)$ and $(0, 1)$.

After that if $\vec{v_1} = x_1 \vec{e_1} + y_1 \vec{e_2}$ and $\vec{v_2} = x_2 \vec{e_2} + y_2 \vec{e_2}$ then by linearity of dot product we have $(\vec{v_1}, \vec{v_2}) = x_1 x_2 (\vec{e_1}, \vec{e_1}) + x_1 y_2 (\vec{e_1}, \vec{e_2}) + x_2 y_1 (\vec{e_2}, \vec{e_1}) + x_2 y_2 (\vec{e_2}, \vec{e_2})$. Since $(\vec{e_1}, \vec{e_1}) = (\vec{e_2}, \vec{e_2}) = 1$ and $(\vec{e_1}, \vec{e_2}) = (\vec{e_2}, \vec{e_1}) = 0$ we have $(\vec{v_1}, \vec{v_2}) = x_1 x_2 + y_1 y_2$.

  • $\begingroup$ How does this show that $v \cdot w = |v| |w| \cos \theta$? $\endgroup$ – Eric O. Korman Aug 1 '10 at 12:33
  • $\begingroup$ As an interesting note, if we are showing the consistency with the inner product, we assume linearity and so we can proceed along the same lines. Of course the theory related to projections takes more work $\endgroup$ – Casebash Aug 1 '10 at 12:37
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    $\begingroup$ @Eric: This answer assumes the definition of dot product as |v||w|cos(θ) and proves that x1x2+y1y2 is the same as it; your answer does it vice-versa. It would be better if both answers made it clearer where they were starting from. (Though from the vaguely stated question, I suspect falagar's answer is slightly closer to the intended spirit.) $\endgroup$ – ShreevatsaR Aug 1 '10 at 12:43
  • $\begingroup$ Ah sorry, I skipped over the part "At first you can prove that dot product is linear..." and thought it just assumed it. $\endgroup$ – Eric O. Korman Aug 1 '10 at 12:49
  • $\begingroup$ @ShreevatsaR: thank you for your comment. I added a line about what I start from. $\endgroup$ – falagar Aug 1 '10 at 15:07

The dot product is invariant under rotations, we may therefore rotate our coordinate system so that v is along the x-axis. In this case, $v = (|v|, 0)$. Letting $w = (x,y)$ we have (using the definition of dot product in Cartesian coordinates) $v \cdot w = |v| x$. But what is $x$? Well, if you draw the picture and let $\theta$ be the angle between v and w, then we see that $\cos \theta = x/|w||$ so that $x = |w| \cos \theta$. Thus $v\cdot w = |v||w| \cos \theta$.

  • $\begingroup$ Is there an easy method to prove invariance under rotations of the x1*x2 + y1*y2 form, without using the equivalence of the two forms? $\endgroup$ – Casebash Aug 1 '10 at 12:41
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    $\begingroup$ It is a straightforward computation once we know the matrix form of rotations. That can be derived pretty easily geometrically by seeing what a rotation does to the basis vectors. $\endgroup$ – Eric O. Korman Aug 1 '10 at 12:53

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