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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