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Question
The torque of a force \[\overrightarrow F \] about a point is defined as \[\overrightarrow\Gamma = \overrightarrow r \times \overrightarrow F.\] Suppose \[\overrightarrow r, \overrightarrow F\] and \[\overrightarrow \Gamma\] are all nonzero. Is \[r \times \overrightarrow\Gamma || \overrightarrow F\] always true? Is it ever true?
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Solution
\[\text{No, }\overrightarrow r \times \overrightarrow \tau || \overrightarrow \Gamma\text{ is not true.} \]
In fact, it is never true.This is because:
\[ \overrightarrow r \times \overrightarrow \tau \]
\[ = \overrightarrow r \times \left( \overrightarrow r \times \overrightarrow F \right)\]
Applying vector triple product, we get:
\[\overrightarrow r \times \left( \overrightarrow r \times \overrightarrow F \right)\]
\[ = \left(\overrightarrow r . \overrightarrow F \right) \overrightarrow r - \left( \overrightarrow r . \overrightarrow r \right) \overrightarrow F \]
\[ \because \overrightarrow r . \overrightarrow r = r^2 \]
\[ = \left( \overrightarrow r . \overrightarrow F \right) \overrightarrow r {}^{-r^2} \overrightarrow F \]
\[\text{If }\overrightarrow r . \overrightarrow F = 0; \text{ that is, }\overrightarrow r {}^{\perp}\overrightarrow F,\text{ then} \]
\[\overrightarrow r \times \overrightarrow \Gamma = {}^{-r^2}\overrightarrow F \]
\[\text{We know that }r^2\text{ is never negative and }\overrightarrow r \times \overrightarrow \Gamma = -r^2 \overrightarrow F \]
This implies that both vectors may be antiparallel to each other but not parallel.
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