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(\right. \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \left.\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 \overset{\rightarrow}{p} = \overset{\rightarrow}{SP} \textrm{ } , \overset{\rightarrow}{q} = \overset{\rightarrow}{SQ} \textrm{ } , \textrm{ } \overset{\rightarrow}{r} = \overset{\rightarrow}{SR} \textrm{ }\textrm{ } and \textrm{ }\textrm{ } \overset{\rightarrow}{t} = \overset{\rightarrow}{ST} then the value of \left|\right. \left(\right. \overset{\rightarrow}{p} \times \overset{\rightarrow}{q} \left.\right) \times \left(\right. \overset{\rightarrow}{r} \times \overset{\rightarrow}{t} \left.\right) \left|\right. is_____

Engineering

Mathematics

Scalar Triple Product

Question

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 $S (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ 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 $\vec{p} = \vec{SP}, \vec{q} = \vec{SQ}, \vec{r} = \vec{SR} and \vec{t} = \vec{ST}$ then the value of $| (\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t}) |$ is______.

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$\vec{SP} = \vec{p} = \frac{\hat{i}}{2} - \frac{\hat{j}}{2} - \frac{\hat{k}}{2};$

$\vec{SQ} = \vec{q} = \frac{\hat{i}}{2} + \frac{\hat{j}}{2} - \frac{\hat{k}}{2};$

$\vec{SR} = \vec{r} = \frac{- \hat{i}}{2} - \frac{\hat{j}}{2} + \frac{\hat{k}}{2};$

$\vec{ST} = \vec{t} = \frac{\hat{i}}{2} + \frac{\hat{j}}{2} + \frac{\hat{k}}{2}$

$= (\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t}) = [\begin{matrix} \vec{p} & \vec{r} & \vec{t} \end{matrix}] \vec{q} - [\begin{matrix} \vec{q} & \vec{r} & \vec{t} \end{matrix}] \vec{p}$

$[\begin{matrix} \vec{p} & \vec{r} & \vec{t} \end{matrix}] = | \begin{matrix} \frac{1}{2} & \frac{- 1}{2} & \frac{- 1}{2} \\ \frac{- 1}{2} & \frac{- 1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{matrix} | = \frac{1}{8} | \begin{matrix} 1 & - 1 & - 1 \\ - 1 & - 1 & 1 \\ 1 & 1 & 1 \end{matrix} | = \frac{1}{8} | \begin{matrix} 0 & - 2 & 0 \\ 0 & 0 & 2 \\ 1 & 1 & 1 \end{matrix} | \frac{1}{8} (- 4) = \frac{- 1}{2}$

$[\begin{matrix} \vec{q} & \vec{r} & \vec{t} \end{matrix}] = | \begin{matrix} \frac{- 1}{2} & \frac{1}{2} & \frac{- 1}{2} \\ \frac{- 1}{2} & \frac{- 1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \end{matrix} | = \frac{1}{8} | \begin{matrix} - 1 & 1 & - 1 \\ - 1 & - 1 & 1 \\ 1 & 1 & 1 \end{matrix} | = \frac{1}{8} | \begin{matrix} - 2 & 0 & 0 \\ 0 & 0 & 2 \\ 1 & 1 & 1 \end{matrix} | = \frac{1}{2}$

$(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t}) = \frac{- \vec{q}}{2} - \frac{\vec{p}}{2} = \frac{- (\vec{p} + \vec{q})}{2} = \frac{- (- \hat{k})}{2} = \frac{\hat{k}}{2} .$

$∴ | (\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t}) | = \frac{1}{2} = 0.50 .$