Scalar Product


The scalar product of two vectors A and B is defined as the product of the magnitudes of A and B and of the cosine of the angle θ formed by A and B:

A ⋅ B = AB cos θ

The scalar product of A and B is a scalar and is denoted by A ⋅ B. The scalar product is also known as the dot product because of its notation.

The scalar product of two vectors is commutative:

A ⋅ B = B ⋅ A

The scalar product is also distributive:

A ⋅ (A1 + A2) = A ⋅ B1 + A ⋅ B2

Elaboration on the distributive relationship:
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Without any loss of generality, it can be assumed that P is directed along the y axis. Denoting by Q the sum of Q1 and Q2 and by θy, the angle Q forms with the y axis, the left–hand member can be expressed as follows:

P·(Q1+Q2);P·Q=PQcos0y=PQy

Where Qy is the y component of Q. In a similar way the right—hand member of (3.26) can be expressed as:

P · Qi + P · Q2 Z P(Qi), + P(Q2)y

Since Q is the sum of Q1 and Q2, its y component must be equal to the sum of the y components of Q1 and Q2. Thus, the expressions obtained in and are equal, and the relation has been proved.

As far as the third property—the associative property—is concerned, we note that this property cannot apply to scalar products. Indeed, (P · Q) - S has no meaning, since P · Q is not a vector but a scalar.

The scalar product of two vectors P and Q can be expressed in terms of their rectangular components. Resolving P and Q into components, we first write

P - Q = (Pxi + Pyj + Pzk) · (Qxi + Qyj + Qzk)

Making use of the distributive property, we express P · Q as the sum of scalar products, such as Pxi · Qxi and Pxi. However, from the definition of the scalar product it follows that the scalar products of the unit vectors are either zero or one.

i·i=l j·j=I k·k=I
i·j;O j·k;O k·i=O

Thus, the expression obtained for P · Q reduces to

P · Q I PxQx + PyQy + PZQZ (1

In the particular case when P and Q are equal, we note that

P·P=Pf+P5+P3=P2 (

Applications

1. Angle formed by two given vectors. Let two vectors be given in terms of their components:

P = Pxi + Pyj + Pzk
Q=Q,i+Qyj+Q;k

To determine the angle formed by the two vectors, we equate the expressions obtained in (3.24) and (3.30) for their scalar product and write

PQ cos 9 Z PxQx + PyQy + PZQZ

Solving for cos 9, we have

Pa X + P1 + P; e cos 0 1 4 Q jQy Q (3 PQ