Question

If a force ] = (+2) -3Â) N is applied at a point whose position vector is r = (-2) +k) m, then unit vector along torque vector about the origin will be i- jak 3 1 / 31 - J - K (ik) 01-1 o i-)- k 13 Toli- j - k)​

Answer 1

Answer:

The unit vector along the torque is $\hat n= \frac{ \hat i + \hat j+ \hat k}{\sqrt{3} }$

Explanation:

The question and the question in figure are both different, I am providing the answer for the image attached, which can be used to solve the given question because the question written is not very clear but can be found the same way.

The force vector is given by $\vec{ F} =\hat i+ 2\hat j-3\hat k$ N

The position vector is given by $\vec{r}=\hat i-2\hat j+ \hat k$ m

The torque at the point  is given b$\tau = \vec{r} \times \vec{F}$, its the cross product of r and F

$\vec{\tau}=(\hat i+ 2\hat j-3\hat k) \times (\hat i-2\hat j+ \hat k)= 4\hat i +4 \hat j+ 4\hat k$

The unit vector along the torque is $\hat n=\frac{\vec{\tau}}{|\tau|}=\frac{ 4\hat i +4 \hat j+ 4\hat k}{4\sqrt{3} }= \frac{ \hat i + \hat j+ \hat k}{\sqrt{3} }$. where $|\tau|$ is the magnitude of torque.