Consider the cube in the first octant with si | vedclass

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(C) The vertices of the cube are $O(0,0,0), P(1,0,0), Q(0,1,0), R(0,0,1)$ and $T(1,1,1)$. The center $S$ is $\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{

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$0.5$

Vedclass · Answer

(C) The vertices of the cube are $O(0,0,0), P(1,0,0), Q(0,1,0), R(0,0,1)$ and $T(1,1,1)$. The center $S$ is $\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}ight)$.Calculating the vectors:$\overrightarrow{p} = \overrightarrow{SP} = (1-\frac{1}{2})\hat{i} + (0-\frac{1}{2})\hat{j} + (0-\frac{1}{2})\hat{k} = \frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}$$\overrightarrow{q} = \overrightarrow{SQ} = (0-\frac{1}{2})\hat{i} + (1-\frac{1}{2})\hat{j} + (0-\frac{1}{2})\hat{k} = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}$$\overrightarrow{r} = \overrightarrow{SR} = (0-\frac{1}{2})\hat{i} + (0-\frac{1}{2})\hat{j} + (1-\frac{1}{2})\hat{k} = -\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}$$\overrightarrow{t} = \overrightarrow{ST} = (1-\frac{1}{2})\hat{i} + (1-\frac{1}{2})\hat{j} + (1-\frac{1}{2})\hat{k} = \frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}$Calculating cross products:$\overrightarrow{p} 	imes \overrightarrow{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1/2 & -1/2 & -1/2 \ -1/2 & 1/2 & -1/2 \end{vmatrix} = \hat{i}(\frac{1}{4} + \frac{1}{4}) - \hat{j}(-\frac{1}{4} - \frac{1}{4}) + \hat{k}(\frac{1}{4} - \frac{1}{4}) = \frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}$$\overrightarrow{r} 	imes \overrightarrow{t} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1/2 & -1/2 & 1/2 \ 1/2 & 1/2 & 1/2 \end{vmatrix} = \hat{i}(-\frac{1}{4} - \frac{1}{4}) - \hat{j}(-\frac{1}{4} - \frac{1}{4}) + \hat{k}(-\frac{1}{4} + \frac{1}{4}) = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}$Now,$|(\overrightarrow{p} 	imes \overrightarrow{q}) 	imes (\overrightarrow{r} 	imes \overrightarrow{t})| = |(\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}) 	imes (-\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j})| = |\frac{1}{4}(\hat{i} + \hat{j}) 	imes (-\hat{i} + \hat{j})| = |\frac{1}{4}(\hat{k} - (-\hat{k}))| = |\frac{1}{4}(2\hat{k})| = |\frac{1}{2}\hat{k}| = \frac{1}{2} = 0.5$

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