Reciprocal vectors

7.9 RECIPROCAL VECTORS
the line to the plane is zero unless
b · n̂ = 0,
in which case the distance, d, will be
d = |(a − r) · n̂|,
where r is any point in the plane.
A line is given by r = a + λb, where a = i + 2j + 3k and b = 4i + 5j + 6k. Find the
coordinates of the point P at 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 that b · n = 0. Thus the line does indeed intersect the plane. To find
the point of intersection we merely substitute the x-, y- and z- 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.
− 14 ,
This gives λ =
which we may substitute into the equation for the line to obtain
x = 1 − 14 (4) = 0, y = 2 − 14 (5) = 34 and z = 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 vectors a, b, c and a , b , c are called reciprocal sets if
a · a = b · b = c · c = 1
(7.47)
a · b = a · c = b · a = b · c = c · a = c · b = 0.
(7.48)
and
It can be verified (see exercise 7.19) that the reciprocal vectors of a, b and c are
given by
b×c
,
a · (b × c)
c×a
,
b =
a · (b × c)
a×b
,
c =
a · (b × c)
a =
(7.49)
(7.50)
(7.51)
where a · (b × c) = 0. In other words, reciprocal vectors only exist if a, b and c are
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