# Question about the dot product [duplicate]

I understand the geometric meaning of cross product but not getting with what dot product of two vectors actually implies geometrically.

• Also related: understanding dot and cross products – JMoravitz Apr 3 '17 at 20:44
• A search option is available on this site for a reason. I suggest you use it in the future. – JMoravitz Apr 3 '17 at 20:44
• It is the area of the rectangle of which one side is one of the vectors and the other is the projection of the other vector on the first. – N74 Apr 3 '17 at 20:47
• By now you should have learnt it. Put related questions where you cannot proceed. – Narasimham Apr 4 '17 at 2:23

The dot product can be defined as:

$$\vec a\cdot\vec b=|\vec a||\vec b| \cos(\theta)$$

Rewritten as:

$$\cos(\theta)=\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$$

$$\theta=\arccos\left(\frac{\vec a\cdot\vec b}{|\vec a||\vec b|}\right)$$

we see that the dot product can be used to calculate the angle between two vectors.

Note if $\theta=\frac{\pi}2$, then the dot product is zero, and the two vectors are orthogonal.

• "We see that the dot product returns the angle between two vectors" ... wouldn't it be much more accurate to say we can use the dot product to compute the angle between to vectors? As stated, it sounds like you are stating $a\cdot b = \theta$ – David Apr 3 '17 at 20:53

Geometrically, the dot product is the product of the magnitudes of the vectors multiplied by the cosine of the angle between them. Since we are only speaking about magnitudes and magnitude is a scalar (has no direction), the dot product is also called the scalar product. The dot product of two vectors is a scalar, it has only magnitude and no direction.

Roughly, if $\textbf x$ is a unit vector, then $\textbf x\cdot\textbf y$ measures the projection of $\textbf y$ in the direction of $\textbf x$ (and by symmetry vice-versa). The actual vector being measured then is $(\textbf x\cdot \textbf y)\textbf x$, since it has the right direction ($\textbf x$) and the right magnitude ($\textbf x\cdot\textbf y)$. I think I said that right.

So in some sense, the dot product and the cross product are complementary notions (though there is something of an asymmetry because one yields a scalar and the other a vector).

• The "so in some sense" part only follows if you're sure OP knows that $\mathbf x\times\mathbf y$ measures the rejection of $\mathbf y$ from the direction of $\mathbf x$. My guess is (s)he does not. Nevertheless, this is the way I prefer to think about the dot product so +1. – user137731 Apr 6 '17 at 17:48
• @Bye_World : In my experience, students learn the cross product as a way to form a vector that is perpendicular to two given vectors, so I would say OP is likely to recognize this. Thanks for the upvote :) – MPW Apr 6 '17 at 18:01