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If
a
′
=
b
×
c
[
a
b
c
]
\mathbf{a}' = \frac{\mathbf{b} \times \mathbf{c}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}
a
′
=
[
abc
]
b
×
c
…
Mathematics
Problem
If
a
′
=
b
×
c
[
a
b
c
]
\mathbf{a}' = \frac{\mathbf{b} \times \mathbf{c}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}
a
′
=
[
abc
]
b
×
c
,
b
′
=
c
×
a
[
a
b
c
]
\mathbf{b}' = \frac{\mathbf{c} \times \mathbf{a}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}
b
′
=
[
abc
]
c
×
a
,
c
′
=
a
×
b
[
a
b
c
]
\mathbf{c}' = \frac{\mathbf{a} \times \mathbf{b}}{[\mathbf{a} \mathbf{b} \mathbf{c}]}
c
′
=
[
abc
]
a
×
b
, then show that
a
×
a
′
+
b
×
b
′
+
c
×
c
′
=
0
,
where
a
,
b
and
c
are non-coplanar.
\mathbf{a} \times \mathbf{a}' + \mathbf{b} \times \mathbf{b}' + \mathbf{c} \times \mathbf{c}' = 0, \text{ where } \mathbf{a}, \mathbf{b} \text{ and } \mathbf{c} \text{ are non-coplanar.}
a
×
a
′
+
b
×
b
′
+
c
×
c
′
=
0
,
where
a
,
b
and
c
are non-coplanar.
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