Page 21 - Intro to Tensor Calculus

17

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In figure 1.1-4 observe that: (i) |B × C| is the area of the parallelogram PQRS. (ii) the unit vector
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B × C
b e n =
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|B × C|
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is normal to the plane containing the vectors B and C. (iii) The dot product

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B × C
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A · b e n = A ·
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|B × C|
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equals the projection of A on b e n which represents the height of the parallelepiped. These results demonstrate
that

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A · (B × C) = |B × C| h = (area of base)(height) = volume.
EXAMPLE 1.1-16. Verify the vector identity

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(A × B) × (C × D)= C(D · A × B) − D(C · A × B)
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Solution: Let F = A × B = F i b e i and E = C × D = E i b e i . These vectors have the components
and
F i = e ijk A j B k E m = e mnp C n D p
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where all indices have the range 1, 2, 3. The vector G = F × E = G i b e i has the components
G q = e qim F i E m = e qim e ijk e mnp A j B k C n D p .
From the identity e qim = e mqi this can be expressed
G q =(e mqi e mnp )e ijk A j B k C n D p
which is now in a form where we can use the e − δ identity applied to the term in parentheses to produce
G q =(δ qn δ ip − δ qp δ in )e ijk A j B k C n D p .

Simplifying this expression we have:

G q = e ijk [(D p δ ip )(C n δ qn )A j B k − (D p δ qp )(C n δ in )A j B k ]
= e ijk [D i C q A j B k − D q C i A j B k ]
= C q [D i e ijk A j B k ] − D q [C i e ijk A j B k ]

which are the vector components of the vector

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C(D · A × B) − D(C · A × B).