Mathematical Methods for Physics and Engineering : A Comprehensive Guide

7.9 RECIPROCAL VECTORS

the line to the plane is zero unless

b·nˆ=0,

in which case the distance,d, will be

d=|(a−r)·ˆn|,

whereris any point in the plane.

A line is given byr=a+λb,wherea=i+2j+3kandb=4i+5j+6k.Findthe

coordinates of the pointPat which the line intersects the plane

x+2y+3z=6.

A vector normal to the plane is

n=i+2j+3k,

from which we find thatb·n= 0. Thus the line does indeed intersect the plane. To find
the point of intersection we merely substitute thex-,y-andz- values of a general point
on the line into the equation of the plane, obtaining

1+4λ+2(2+5λ)+3(3+6λ)=6 ⇒ 14 + 32λ=6.

This givesλ=−^14 , which we may substitute into the equation for the line to obtain
x=1−^14 (4) = 0,y=2−^14 (5) =^34 andz=3−^14 (6) =^32. Thus the point of intersection is

(0,^34 ,^32 ).

7.9 Reciprocal vectors

The final section of this chapter introduces the concept of reciprocal vectors,

which have particular uses in crystallography.

The two sets of vectorsa,b,canda′,b′,c′are calledreciprocal setsif

a·a′=b·b′=c·c′= 1 (7.47)

and

a′·b=a′·c=b′·a=b′·c=c′·a=c′·b=0. (7.48)

It can be verified (see exercise 7.19) that the reciprocal vectors ofa,bandcare

given by

a′=

b×c

a·(b×c)

, (7.49)

b′=

c×a

a·(b×c)

, (7.50)

c′=

a×b

a·(b×c)

, (7.51)

wherea·(b×c)= 0. In other words, reciprocal vectors only exist ifa,bandcare