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Consider two vectors $V_1$ and $V_2$ in $\mathbb{R}^3$. When we take their dot product we get a real number. How is that number related to the vectors? Is there any way we can visualize it?

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    $\begingroup$ $u \cdot v = \|u\| \|v\| \cos \theta$ $\endgroup$
    – user61527
    Jul 3, 2014 at 7:48
  • $\begingroup$ @T.Bongers With one or two more words, that could be an answer... $\endgroup$
    – user98602
    Jul 3, 2014 at 8:04

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Temporarily imagine that $V_2$ is of unit length. Then, $V_1 \cdot V_2$ is the projection of the vector $V_1$ onto the vector $V_2$. Picture here. Now we let $V_2$ have its original length and to do so we multiply the result of the dot product by the new length of $V_2$. (This has the effect of making it not matter which one you pretend has unit length initially.)

You do this sort of thing when you write a vector as a sum of multiples of the standard unit coordinate vectors (sometimes written $\hat{x}, \hat{y}$, and $\hat{z}$). Use the dot product to project your vector onto $\hat{x}$ getting the multiple of $\hat{x}$ that, when assembled with the other components will sum to your vector.

The dot product is a (poor) measure of the degree of parallelism of two vectors. If they point in the same (or opposite) directions, then the projection of one onto the other is not just a component of the length of the projected vector, but is the entire projected vector. It is a poor measure because it is scaled by the lengths of the two vectors -- so one has to know not only their dot product, but also their lengths, to determine how parallel or perpendicular they really are.

In physics,the dot product is frequently used to determine how parallel some vector quantity is to some geometric construct, for instance the direction of motion (displacement) versus a partially opposing force (to find out how much work must be expended to overcome the force). Another example is the direction of the electric field compared to a small patch of surface (which is represented by a vector "normal" to its surface and of length proportional to its area).

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  • $\begingroup$ If $V_1 \cdot V_2$ is the projection of $V_1$ onto $V_2$, then the reverse is also true right, i.e., $V_1 \cdot V_2 = V_2 \cdot V_1$ is the orthogonal project of $V_2$ onto $V_1$? $\endgroup$
    – David
    Jun 4, 2020 at 20:26
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    $\begingroup$ @David : Yes, the dot product is commutative. I said as much: "(This has the effect of making it not matter which one you pretend has unit length initially.)" $\endgroup$ Jun 4, 2020 at 20:49
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The dot product ${\bf a}\cdot{\bf b}$ measures the length of ${\bf a}$'s orthogonal projection onto $\bf b$ (the $1$-dimensional subspace it is a part of), scaled by the length of $\bf b$ itself. And conversely. The scaling is nice to have because it means the dot product is bilinear in its two arguments. Physically, it can measure how much of something in one direction moves in a different direction. For instance, suppose water is moving through a net (in the shape of a plane) put in the ocean - and, for simplicity, the water is moving monolithically: the same direction at the same speed at every point - but the direction of movement is not perfectly perpendicular to the net. The dot product of the net's unit normal vector with the water's velocity vector will tell us how much is moving through the net. When this is applied on an infinitessimal scale and then integrated, we can determine how much of something moving with a vector field is displaced through a surface; this is half of the divergence theorem.

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