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Vectors Help.

Started by As-12 Apr 9, 2006 at 3:56 AM 3 replies 550+ views
Original Post
As-12
As-12
Hmm... Im studying vectors , dot product and I dont get it.Well Dot product return a scalar value right? So this scalar is ? What it use for? like we have 2 vectors and its between angle = 90 degree. So the dot product = 0 right? What is 0 use for? Im studying from http://chortle.ccsu.edu/VectorLessons/vectorIndex.html Sry if my grammar suck. Im asian and don't speak english .
Zakwayda
Zakwayda
Any good reference on vectors will tell you that the value returned by the dot product is the cosine of the angle between the two vectors, scaled by the length of both:
dot(a,b) = cos(theta)*|a|*|b|
The dot product has other properties as well, such as:
dot(v,v) = the squared length of vdot(a,b) = the length of the projection of a onto b scaled by |b| and vice versadot(a,b) = cosine(theta) when |a| = |b| = 1dot(a,b) = 0 when a and b are perpendiculardot(a,b) > 0 when a and b are pointing more in the same direction than oppositedot(a,b) < 0 when a and b are pointing more in opposite directions than the same
And so on. Spending some time working with vectors will make the properties and usefulness of the dot product more clear.
etothex
etothex
In vector spaces (well, for the nitpicky, Hilbert spaces) the dot product is also called the "inner" product and it is the base on which such fundamental principles as "distance" and "angle" are defined. Otherwise, it is too difficult to define such abstract ideas rigorously.

In the usual "euclidean" space that we think of, then the dot product relates to distance and angle in the ways jky mentioned.
Sheep
Sheep
The dot product is a measure of orthogonality of two vectors. The dot product appears extensively in physics. Perhaps the simplest mechanical example is work. For constant displacement vector d and force F, the work done moving an object along the displacement vector is where <.,.> is the dot product. If we push an object along a horizontal plane, we do zero work because gravitational force is perpendicular to the motion, so = |F| |d| cos 90 = 0.

The inner product even defines orthogonality on an arbitrary inner product space (ethotex - not neccessarily a Hilbert space, it doesn't have to be complete) where there is no obvious geometric interpretation, and distance can be defined with the natural norm sqrt().
etothex
etothex
Quote:
Original post by Sheep

The inner product even defines orthogonality on an arbitrary inner product space (ethotex - not neccessarily a Hilbert space, it doesn't have to be complete) where there is no obvious geometric interpretation, and distance can be defined with the natural norm sqrt().


True, though since any inner product space has a metric derived from the inner product, it can be naturally completed into a proper Hilbert space, and it is often useful to think of the derived Hilbert spaces rather than any random inner product space. But you're right, of course.

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