Consider the cube in the first octant with sides $OP, OQ$ and OR of length $1$, along the x-axis, y-axis and z-axis, respectively, where $O(0,0,0)$ is the origin. Let $S\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$ be the centre of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal $OT$. If $\overrightarrow{p}=\overrightarrow{SP}, \overrightarrow{q}=\overrightarrow{SQ}, \overrightarrow{r}=\overrightarrow{SR}$ and $\overrightarrow{t}=\overrightarrow{ST}$, then the value of $\left|\left(\overrightarrow{p}\times \overrightarrow{q}\right)\times \left(\overrightarrow{r}\times \overrightarrow{t}\right)\right|$ is ______.

Solution

$\overrightarrow{p}=\overrightarrow{SP}=\left(\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\right)=\frac{1}{2}\left(\widehat{i}-\widehat{j}-\widehat{k}\right)$
$\overrightarrow{q}=\overrightarrow{SQ}=\left(-\frac{1}{2},\frac{1}{2},-\frac{1}{2}\right)=\frac{1}{2}\left(-\widehat{i}+\widehat{j}-\widehat{k}\right)$
$\overrightarrow{r}=\overrightarrow{SR}=\left(-\frac{1}{2},-\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2}\left(-\widehat{i}-\widehat{j}+\widehat{k}\right)$
$\overrightarrow{t}=\overrightarrow{ST}=\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)=\frac{1}{2}\left(\widehat{i}+\widehat{j}+\widehat{k}\right)$
$\left|\left(\overrightarrow{p}\times \overrightarrow{q}\right)\times \left(\overrightarrow{r}\times \overrightarrow{t}\right)\right|=\left|\frac{1}{4}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& -1\\ -1& 1& -1\end{array}\right|\times \frac{1}{4}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& -1& 1\\ 1& 1& 1\end{array}\right|\right|$
$=\frac{1}{16}\left|\left(2\widehat{i}+2\widehat{j}\right)\times \left(-2\widehat{i}+2\widehat{j}\right)\right|=\left|\frac{\widehat{k}}{2}\right|=\frac{1}{2}$

Hence, the correct answer is 0.5.