Let a force $\overrightarrow{\mathrm{F}}=-\mathrm{F} \mathrm{k}$ acts on the origin of cartesian frame of…
Let a force $\overrightarrow{\mathrm{F}}=-\mathrm{F} \mathrm{k}$ acts on the origin of cartesian frame of reference. The moment of force about a point $(1,-1)$ will be
$-\mathrm{F}(\hat{\imath}+\hat{\mathrm{\jmath}})$
$-\mathrm{F}(\hat{\mathrm{\imath}}-\hat{\jmath})$
$\mathrm{F}(\hat{\imath}-\hat{\jmath})$
$\mathrm{F}(\hat{\imath}+\hat{\jmath})$
Solution
The point $\mathrm{P}$ lies at $(1,-1)$ i.e, $\overrightarrow{\mathrm{P}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}$
$\Rightarrow \overrightarrow{\mathrm{PO}}=\overrightarrow{\mathrm{O}}-\overrightarrow{\mathrm{P}}=-\hat{\mathrm{P}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}$
Torque about point $\mathrm{P}$:
$\Rightarrow \vec{\tau}=\overrightarrow{\mathrm{PO}} \times \overrightarrow{\mathrm{F}}=(-\hat{\mathrm{i}}+\hat{\mathrm{j}}) \times(-\mathrm{F} \hat{\mathrm{k}})$
$\Longrightarrow \vec{\tau}=\mathrm{F}[(\hat{\mathrm{i}} \times \hat{\mathrm{k}})+\hat{\mathrm{k}} \times \hat{\mathrm{j}}]$
Using: $\hat{\mathrm{i}} \times \hat{\mathrm{j}}=\hat{\mathrm{k}} ; \hat{\mathrm{j}} \times \hat{\mathrm{k}}=\hat{\mathrm{i}} ; \hat{\mathrm{k}} \times \hat{\mathrm{i}}=\hat{\mathrm{j}}$
$\Rightarrow \vec{\tau}=\mathrm{F}(-\hat{\mathrm{i}}-\hat{\mathrm{j}})=-\mathrm{F}(\hat{\mathrm{i}}+\hat{\mathrm{j}})$