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?

Answer 1

\[\text{No, }\overrightarrow r  \times  \overrightarrow \tau ||  \overrightarrow \Gamma\text{ is not true.} \] 

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.