Mathematics · Vector Algebra

JEE Advanced 2018 — Paper 2 — Question 40

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 $\mathrm{O}(0,0,0)$ 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 $\vec{p}=\overrightarrow{S P}, \vec{q}=\overrightarrow{S Q}, \vec{r}=\overrightarrow{S R}$ and

$\vec{t}=\overrightarrow{S T}$ , then the value of $|(\vec{p} \times \vec{q}) \times(\vec{r} \times \vec{t})|$ is $\_\_\_\_\_$ .

Answer: 0.5

Numerical answer — enter this value.

Step-by-step solution

Let the vertices be $O(0,0,0), P(1,0,0), Q(0,1,0), R(0,0,1), T(1,1,1)$ . The centre is $S(\frac12,\frac12,\frac12)$ . Compute vectors: $\vec{p} = \overrightarrow{SP} = (\frac12, -\frac12, -\frac12)$ , $\vec{q} = \overrightarrow{SQ} = (-\frac12, \frac12, -\frac12)$ , $\vec{r} = \overrightarrow{SR} = (-\frac12, -\frac12, \frac12)$ , $\vec{t} = \overrightarrow{ST} = (\frac12, \frac12, \frac12)$ . Compute $\vec{p} \times \vec{q} = (\frac12, \frac12, 0)$ . Compute $\vec{r} \times \vec{t} = (-\frac12, \frac12, 0)$ . Now $(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t}) = (0, 0, \frac12)$ . Magnitude: $|(0,0,\frac12)| = \frac12 = 0.5$ . (Alternatively, using the identity $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = [\vec{a}\vec{c}\vec{d}]\vec{b} - [\vec{b}\vec{c}\vec{d}]\vec{a}$ , with $[\vec{p}\vec{r}\vec{t}] = -\frac14$ , $[\vec{q}\vec{r}\vec{t}] = \frac14$ , leading to the same result.)

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Mathematics
Chapter: Vector Algebra
Topic: Vector or Cross Product of Two Vectors