Tuesday, February 5, 2013

Dot product


There are two types of multiplication of two vectors. That is, as two types of products. One of them gives the result as a vector quantity and the other results as a scalar quantity. They are termed as cross product and dot product respectively.  The names are given so because in the first case a ‘x’ is used to indicate the multiplication and in the second case a ‘•’ indicates the same operation. Because of the fact that a dot product results in scalar quantity, the dot-product is also called as scalar dot product.

If P and Q are two vectors, then the dot-product two vectors is denoted as P • Q and its magnitude is given by lPl*lQlcos θ, where θ is the angle between the vectors P and Q. Unlike vector products, scalar products of vectors are commutative.

A typical example of a scalar product is the calculation of work done W by a force F while displacing the object by a displacement D. As per physics concept, W = lFl*lDlcos θ, where θ is the angle between the force and the direction of displacement. Although, F and D are vector quantities, the work done W is a scalar quantity. The work done by a force has no direction. .

From the definition, P • Q = lPl*lQlcos θ, let us analyze when the dot product is zero. It is obvious that it happens when P or Q or both are 0. But there is one more possibility. The scalar product becomes 0 when the angle between the vectors is 90o, even though P and Q may have finite values, since cos 90o = 0. In other words, the scalar product of two vectors acting at right angle is 0. We can illustrate this fact with a practical example. We all know that the weight W of an object placed on a smooth horizontal surface is a vector acting vertically. Suppose the object is moved horizontally by a displacement D, the vectors W and D act at right angle. Since the scalar product is 0, the work done W is 0. That is, no work is needed to be done in moving an object at a horizontal level subject to the assumption that there is no friction between the object and the horizontal surface. That is why we find that only a little amount of effort is needed to push even a heavy object horizontally. It is not zero as per theory, because the assumption of ‘no friction’ is practically impossible and the little effort only accounts for work done against the frictional force.

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