VECTOR ALGEBRA
From (7.15) we see that the scalar product has the particularly useful propertythat
a·b= 0 (7.16)is a necessary and sufficient condition forato be perpendicular tob(unless either
of them is zero). It should be noted in particular that the Cartesian basis vectors
i,jandk, being mutually orthogonal unit vectors, satisfy the equations
i·i=j·j=k·k=1, (7.17)i·j=j·k=k·i=0. (7.18)Examples of scalar products arise naturally throughout physics and in partic-ular in connection with energy. Perhaps the simplest is the work doneF·rin
moving the point of application of a constant forceFthrough a displacementr;
notice that, as expected, if the displacement is perpendicular to the direction of
the force thenF·r= 0 and no work is done. A second simple example is afforded
by the potential energy−m·Bof a magnetic dipole, represented in strength and
orientation by a vectorm, placed in an external magnetic fieldB.
As the name implies, the scalar product has a magnitude but no direction. Thescalar product is commutative and distributive over addition:
a·b=b·a (7.19)a·(b+c)=a·b+a·c. (7.20)Four non-coplanar pointsA, B, C, Dare positioned such that the lineADis perpendicular
toBCandBDis perpendicular toAC. Show thatCDis perpendicular toAB.Denote the four position vectors bya,b,c,d. As none of the three pairs of lines actually
intersect, it is difficult to indicate their orthogonality in the diagram we would normally
draw. However, the orthogonality can be expressed in vector form and we start by noting
that, sinceAD⊥BC, it follows from (7.16) that
(d−a)·(c−b)=0.Similarly, sinceBD⊥AC,
(d−b)·(c−a)=0.Combining these two equations we find
(d−a)·(c−b)=(d−b)·(c−a),which, on mutliplying out the parentheses, gives
d·c−a·c−d·b+a·b=d·c−b·c−d·a+b·a.Cancelling terms that appear on both sides and rearranging yields
d·b−d·a−c·b+c·a=0,which simplifies to give
(d−c)·(b−a)=0.From (7.16), we see that this implies thatCDis perpendicular toAB.