$(1, 1) \cdot (6, 0) = 6?$ Intuition?

Question

$a = (1, 1)$

$b= (6, 0)$

$a \cdot b = (1, 1) \cdot (6, 0) = 6$

I have seen the dot product of $a$ and $b$ refered to as "What is the x-coordinate of $a$, assuming $b$ is the $x$-axis?".

Well here $b$ is actually the $x$-axis so I would have expected the $x$-co-ordinate of $a$ in this case to be $1$, but it's not...What is the intuition behind it being $6$?

Answer 1

Hint: (1,1) dot (1,0) = 1.

What does replacing (1,0) by (6,0) do?

Answer 2

Dot product gives you not a simple projection of $a$ on $b$, but the signed length $l$ of vector $c$ lying on $b$, such that $||c|| = ||a|| * ||b|| * |cos(angle(a, b))|$ and $angle(a, c) < \pi/2$. I.e, |l| = ||c||, $sign(l) = sign(cos(angle(a, b)))$

Answer 3

The dot product can be defined $$a\cdot b =(a_1,a_2)\cdot(b_1,b_2) = a_1b_1+a_2b_2.$$ Why do we bother to define this? Well, it's very useful. For example, it can be shown that $$\cos\angle ab = \frac{a\cdot b}{\mid a\mid\mid b\mid}.$$ That's really useful to me. I can find the angle between two vectors by using the dot product. Let's take a look at the dot product a little further:

$$a\cdot b = (a_1,a_2)\cdot(b_1,b_2)=(a_1,a_2)\cdot(b_1,0)+(a_1,a_2)\cdot(0,b_2).$$

But we can write this as

$$(a_1,a_2)\cdot(1,0)b_1+(a_1,a_2)\cdot(0,1)b_2.$$

So as you can see the first dot product, $$(a_1,a_2)\cdot(1,0)b_1,$$ is the projection of $(a_1,a_2)$ onto the $x$-axis multiplied by the constant $b_1$. Similar for the right hand dot product.

Hope this helps a bit.

Answer 4

In physics ... you have a force $(1,1)$ acting on a body, which moves (constrained to a track on the $x$-axis, say) from $(0,0)$ to $(6,0)$. What is the work done? Answer: the dot product $(1,1) \cdot (6,0) = 6$. So for the intuition, being constrained to move on the $x$-axis, only the first component of the force contributes to this motion. The perpendicular component doesn't make it move.