JEE Advanced 2018 · Paper 2 · Mathematics Q13

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MathematicsNumerical+3Question 49 of 54

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 $\vec{p} = \overrightarrow{SP}$ , $\vec{q} = \overrightarrow{SQ}$ , $\vec{r} = \overrightarrow{SR}$ and $\vec{t} = \overrightarrow{ST}$ , then the value of $|(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t})|$ is _____ .

Your answer

Answer and worked solution

Official answer: 0.5

Vectors from the centre $S(\tfrac12,\tfrac12,\tfrac12)$ :
$\vec p=\tfrac12(1,-1,-1),\quad\vec q=\tfrac12(-1,1,-1),\quad\vec r=\tfrac12(-1,-1,1),\quad\vec t=\tfrac12(1,1,1).$

Cross products.
$\vec p\times\vec q=\tfrac14(2,2,0)=\left(\tfrac12,\tfrac12,0\right),\qquad\vec r\times\vec t=\tfrac14(-2,2,0)=\left(-\tfrac12,\tfrac12,0\right).$

Their cross product.
$\left(\tfrac12,\tfrac12,0\right)\times\left(-\tfrac12,\tfrac12,0\right)=\left(0,0,\tfrac14+\tfrac14\right)=\left(0,0,\tfrac12\right).$
The magnitude is $\tfrac12$ .

Answer: 0.5

Topics:Cross product

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JEE Advanced 2018 · Paper 2, Mathematics question 13