Question
Find the torque of a force $$F = - 3\hat i + \hat j + 5\hat k$$ acting at the point $$r = 7\hat i + 3\hat j + \hat k.$$
A.
$$ - 21\hat i + 3\hat j + 5\hat k$$
B.
$$ - 14\hat i + 3\hat j - 16\hat k$$
C.
$$4\hat i + 4\hat j + 6\hat k$$
D.
$$14\hat i - 38\hat j + 16\hat k$$
Answer :
$$14\hat i - 38\hat j + 16\hat k$$
Solution :
Given, $$r = 7\hat i + 3\hat j + \hat k,F = - 3\hat i + \hat j + 5\hat k$$
$$\therefore \tau = r \times F = \left| r \right|\left| F \right|\sin \theta $$
where, $$\theta $$ is the angle between $$r$$ and $$F$$
$$ = \left( {7\hat i + 3\hat j + \hat k} \right) \times \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
\[ = \left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
7&3&1\\
{ - 3}&1&5
\end{array}} \right| = \hat i\left( {15 - 1} \right) - \hat j\left( {35 + 3} \right) + \hat k\left( {7 + 9} \right)\]
$$ = 14\hat i - 38\hat j + 16\hat k$$
Alternative
$$\eqalign{
& \therefore \tau = r \times F \cr
& = \left( {7\hat i + 3\hat j + \hat k} \right) \times \left( { - 3\hat i + \hat j + 5\hat k} \right) \cr
& = - 21\left( {\hat i \times \hat i} \right) + 7\left( {\hat i \times \hat j} \right) + 35\left( {\hat i \times \hat k} \right) - 9\left( {\hat j \times \hat i} \right) + 3\left( {\hat j \times \hat j} \right) + 15\left( {\hat j \times \hat k} \right) - 3\left( {\hat k \times \hat i} \right) + \left( {\hat k \times \hat j} \right) + 5\left( {\hat k \times \hat k} \right) \cr
& = 0 + 7\hat k - 35\hat j + 9\hat k + 0 + 15\hat i - 3\hat j - 1 + 0 \cr
& = 14\hat i - 38\hat j + 16\hat k \cr} $$