# How does mathematicians reconcil the fact that the gradient not drawn with respect to the origin

For example, consider the gradient vector of a function

Obviously, $\nabla f(x_o,y_o)$ on the above figure is not attached to the origin. But attached to the point for which it is evaluated at.

But shouldn't it be attached to the origin instead? Isn't that the correct interpretation of a vector on $\mathbb{R}^2$

You know? Just the way we learned it in high school?

• No. Vectors are all relative movement, not absolute position. Otherwise, it would be impossible to add two vectors together graphically (like in your picture). Commented Feb 29, 2016 at 1:41
• When you think of a vector, you can visualize an arrow ("directed line segment"), but you have to remember that if two different arrows have the same magnitude and direction then they represent the same vector. There's nothing special about the arrow that's based at the origin. Commented Feb 29, 2016 at 1:46
• @littleQ But given any vector $\vec v$, it is the same as $\vec v$ = $\vec v - 0$, so all vectors are attached to the origin. Commented Feb 29, 2016 at 1:49
• @TheSilenceoftheCows Actually, $\vec v - \vec v = \vec 0$, the zero vector, not the zero point. Commented Feb 29, 2016 at 1:55
• No, at the risk of using undefined terms, I'd say that a vector is not "attached" to anything. A vector has a magnitude and a direction, but it does not have a position. One definition of a vector is an "equivalence class" of directed line segments. Two different arrows with the same magnitude and direction represent the same vector. Commented Feb 29, 2016 at 1:56

A vector from point $p_1=(x_1,y_1)$ to point $p_2=(x_2,y_2)$ can be found by subtracting the corresponding coordinates of $p_1$ from $p_2$. That is,

$$\vec{p_1p_2} = \langle x_2-x_1, y_2-y_1\rangle$$

This follows from the fact that we move $x_2-x_1$ units on the $x$-axis from $x_1$ to get to $x_2$, and we move $y_2-y_1$ units on the $y$-axis from $y_1$ to get to get $y_2$.

As an example, consider the points $(1,2)$ and $(4,6)$. Then, the vector between them would be $\langle 3,4 \rangle$, because we moved three units to the right of $1$ and up $4$ units from $2$.

But now consider the points $(10,11)$ and $(13,16)$. The vector between these points is also $\langle 3, 4 \rangle$.

Now consider the points $(0,0)$ and $(3,4)$. The vector between these points is $\langle 3,4 \rangle$.

As mentioned in the comments, vectors only have magnitudes and directions. They don't have a position because it doesn't matter where we start our vector -- we'd always get the same vector, like above.

• I was about ready to post that the difference of two points is a vector. Your explanation is the next step after that. Commented Feb 29, 2016 at 2:26