JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 13)
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 $$S\left( {{1 \over 2},{1 \over 2},{1 \over 2}} \right)$$ 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 p = SP, q = SQ, r = SR and t = ST, then the value of |(p $$ \times $$ q) $$ \times $$ (r $$ \times $$ t)| is ............
Answer
0.5
Explanation
Here, P(1, 0, 0), Q(0, 1, 0), R(0, 0, 1), T = (1, 1, 1) and $$S\left( {{1 \over 2},{1 \over 2},{1 \over 2}} \right)$$.
Image
Now, $$\overrightarrow p = \overrightarrow {SP} = \overrightarrow {OP} - \overrightarrow {OS} $$
$$ = \left( {{1 \over 2}\widehat i - {1 \over 2}\widehat j - {1 \over 2}\widehat k} \right) = {1 \over 2}(\widehat i - \widehat j - \widehat k)$$
$$\overrightarrow q = \overrightarrow {SQ} = {1 \over 2}( - \widehat i + \widehat j - \widehat k)$$
$$\overrightarrow r = \overrightarrow {SR} = {1 \over 2}( - \widehat i - \widehat j + \widehat k)$$
and $$\overrightarrow t = \overrightarrow {ST} = {1 \over 2}(\widehat i + \widehat j + \widehat k)$$
$$\overrightarrow p \times \overrightarrow q = {1 \over 4}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 1} & { - 1} \cr { - 1} & 1 & { - 1} \cr } } \right| = {1 \over 4}(2\widehat i + 2\widehat j)$$
and $$\overrightarrow r \, \times \,\overrightarrow t = {1 \over 4}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & { - 1} & 1 \cr 1 & 1 & 1 \cr } } \right| = {1 \over 4}( - 2\widehat i + 2\widehat j)$$
Now, $$(\overrightarrow p \, \times \,\overrightarrow q )\, \times \,(\overrightarrow r \, \times \,\overrightarrow t ) = {1 \over {16}}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 2 & 0 \cr { - 2} & 2 & 0 \cr } } \right| = {1 \over {16}}(8\widehat k) = {1 \over 2}\widehat k$$
$$ \therefore $$ $$|(p \times q) \times (\overrightarrow r \, \times \,\overrightarrow t )|\, = \,\left| {{1 \over 2}\widehat k} \right| = {1 \over 2} = 0.5$$
Image
Now, $$\overrightarrow p = \overrightarrow {SP} = \overrightarrow {OP} - \overrightarrow {OS} $$
$$ = \left( {{1 \over 2}\widehat i - {1 \over 2}\widehat j - {1 \over 2}\widehat k} \right) = {1 \over 2}(\widehat i - \widehat j - \widehat k)$$
$$\overrightarrow q = \overrightarrow {SQ} = {1 \over 2}( - \widehat i + \widehat j - \widehat k)$$
$$\overrightarrow r = \overrightarrow {SR} = {1 \over 2}( - \widehat i - \widehat j + \widehat k)$$
and $$\overrightarrow t = \overrightarrow {ST} = {1 \over 2}(\widehat i + \widehat j + \widehat k)$$
$$\overrightarrow p \times \overrightarrow q = {1 \over 4}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 1} & { - 1} \cr { - 1} & 1 & { - 1} \cr } } \right| = {1 \over 4}(2\widehat i + 2\widehat j)$$
and $$\overrightarrow r \, \times \,\overrightarrow t = {1 \over 4}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & { - 1} & 1 \cr 1 & 1 & 1 \cr } } \right| = {1 \over 4}( - 2\widehat i + 2\widehat j)$$
Now, $$(\overrightarrow p \, \times \,\overrightarrow q )\, \times \,(\overrightarrow r \, \times \,\overrightarrow t ) = {1 \over {16}}\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 2 & 0 \cr { - 2} & 2 & 0 \cr } } \right| = {1 \over {16}}(8\widehat k) = {1 \over 2}\widehat k$$
$$ \therefore $$ $$|(p \times q) \times (\overrightarrow r \, \times \,\overrightarrow t )|\, = \,\left| {{1 \over 2}\widehat k} \right| = {1 \over 2} = 0.5$$
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