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?

  • 1
    $\begingroup$ 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, 2010 at 13:25

2 Answers 2


I suppose you want to prove that two your definitions of dot product are the same. We start with definition of dot product as $(\vec{u}, \vec{v}) = |\vec{u}| |\vec{v}| \cos \theta$ 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 $v \cdot w = |v| |w| \cos \theta$? $\endgroup$ Aug 1, 2010 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, 2010 at 12:37
  • 5
    $\begingroup$ @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.) $\endgroup$ Aug 1, 2010 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. $\endgroup$ Aug 1, 2010 at 12:49
  • $\begingroup$ @ShreevatsaR: thank you for your comment. I added a line about what I start from. $\endgroup$
    – falagar
    Aug 1, 2010 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, 2010 at 12:41
  • 1
    $\begingroup$ 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. $\endgroup$ Aug 1, 2010 at 12:53
  • $\begingroup$ If $Q$ is a an orthogonal matrix, then $Q^{T} Q = I$, where $I$ the identidy. Hence, $(Qy)^{T} (Qx) = y^T Q^{T} Qx = y^{T} x$, then the dot product is invariant to orthogonal matrix. It's only left to us to find a such a $Q$ matrix, which it can be choosen to be a Householder trasnformation as done here: en.wikipedia.org/wiki/Householder_transformation. In 2D, it can be choosen to be a Givens rotations as well as done here en.wikipedia.org/wiki/Givens_rotation. P.S.: I vote to make everyone drop the term rotations in favor to orthogonal transformations; this just confuses all $\endgroup$ Sep 7, 2021 at 22:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .