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Thus, the cube has a volume of
1 units.
Now let's interchange the base vectors so that
+
a
=−
i
b
=−
k
c
=
j
which are organised in a left-hand sequence.
Substituting these in Eq. (2.16), we obtain
abc
=
i
×−
k
·
j
Let's compute the cross product:
×−
i
k
i
j k
i
×−
k
=
100
00
=−
j
1
Now compute the dot product :
j
·
j
=−
1 units
Thus, the cube has a volume of
1 units, and we see that
abc
=
Volume
=
a
×
b
·
c
=
a
·
b
×
c
(2.17)
We can compact the above calculation as follows: given
a
=
x a i
+
y a j
+
z a k
b
=
x b i
+
y b j
+
z b k
=
+
+
c
x c i
y c j
z c k
then
i
j k
a
·
b
×
c
=
a
·
x b
y b
z b
x c
y c
z c
y b z c
y c z b i
+
a
·
b
×
c
=
x a i
+
y a j
+
z a k
·
z b x c
z c x b j
+
x b y c
x c y b k
and
a
·
b
×
c
=
x a y b z c
y c z b
+
y a z b x c
z c x b
+
z a x b y c
x c y b
which equals
x a
y a
z a
·
×
=
x b
y b
z b
a
b
c
(2.18)
x c
y c
z c
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