What does the scalar product of two displacements represent?

[andylatham82]

TL;DR

The scalar product of two displacements gives a displacement squared (area). What does this represent?

Hi,

This feels like such a stupid question, but it's bugging me. Two displacements can be represented with two vectors. Let's say their magnitudes are expressed in metres. The scalar (dot) product of the two vectors results in a value with the units of square metres, which must be an area. Can anyone tell me what this area is in relation to the two vectors?

Many thanks!

 

[berkeman]

Summary:: The scalar product of two displacements gives a displacement squared (area). What does this represent?
Two displacements can be represented with two vectors. Let's say their magnitudes are expressed in metres. The scalar (dot) product of the two vectors results in a value with the units of square metres, which must be an area. Can anyone tell me what this area is in relation to the two vectors?

From my undergrad alma mater...

https://www.math.ucdavis.edu/~dadde...plications/Determinant/Determinant/node4.html

 

[A.T.]

Can anyone tell me what this area is in relation to the two vectors?

https://en.wikipedia.org/wiki/Dot_product#Geometric_definition

You can interpret this as the area of a rectangle, with the sides:
- length of vector1
- length of vector2 projection onto vector1

But note that this is a signed area, which goes negative if the angle is > 90°.

 

[A.T.]

From my undergrad alma mater...

https://www.math.ucdavis.edu/~dadde...plications/Determinant/Determinant/node4.html

View attachment 256982

This seems to be about the cross product.

 

[berkeman]

This seems to be about the cross product.

Dagnabit! I blame Google, and my alma mater, of course.

Thanks!

 

[andylatham82]

https://en.wikipedia.org/wiki/Dot_product#Geometric_definition

You can interpret this as the area of a rectangle, with the sides:
- length of vector1
- length of vector2 projection onto vector1

But note that this is a signed area, which goes negative if the angle is > 90°.

Thanks very much for the response. So the resultant area doesn't really represent anything tangible with regards to the two displacements? It's just an abstract 'area' that you couldn't, say, draw on a diagram of the two vectors?

 

[andylatham82]

Dagnabit! I blame Google, and my alma mater, of course.

Thanks!

Thanks for the link anyway, I was also wondering what the cross product would represent, so you've answered that question for me!

 

[A.T.]

It's just an abstract 'area' that you couldn't, say, draw on a diagram of the two vectors?

Nothing stops you from drawing that rectangle torgether with the vectors and their projection.

 

[kuruman]

Thanks for the link anyway, I was also wondering what the cross product would represent, so you've answered that question for me!

To complete the picture, the scalar product of the cross product with a third vector $\vec c$ is the volume of the parallelepiped defined by the three vectors,$$V=(\vec a \times \vec b)\cdot \vec c=(\vec c \times \vec a)\cdot \vec b=(\vec b \times \vec c)\cdot \vec a.$$Note that the third vector must be in the direction of the cross product to ensure a positive volume. If the three vectors are coplanar, the volume is, obviously, zero.