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3D Geometry question

2018 · Shift 2 · Q31
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3D Geometry question

2018 · Shift 2 · Q31

JEE AdvancedMathematics3D GeometryNumerical+3 / −1
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(12,12,12)S\left( {{1 \over 2},{1 \over 2},{1 \over 2}} \right)S(21​,21​,21​) 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 ............
Numerical answer
View written solutionFree

Correct answer: 0.5

  1. Assign coordinates to the cube vertices

Since the cube has side length 111 and edges along the coordinate axes:

  • O=(0,0,0)O=(0,0,0)O=(0,0,0)
  • P=(1,0,0)P=(1,0,0)P=(1,0,0)
  • Q=(0,1,0)Q=(0,1,0)Q=(0,1,0)
  • R=(0,0,1)R=(0,0,1)R=(0,0,1)
  • The opposite vertex is T=(1,1,1)T=(1,1,1)T=(1,1,1)
  • The centre is given as S=(12,12,12)S=\left(\tfrac12,\tfrac12,\tfrac12\right)S=(21​,21​,21​)
  1. Form the vectors

Given p=SP→,q=SQ→,r=SR→,t=ST→p=\overrightarrow{SP},\quad q=\overrightarrow{SQ},\quad r=\overrightarrow{SR},\quad t=\overrightarrow{ST}p=SP,q=SQ​,r=SR,t=ST

So, p=P−S=(12,−12,−12)p=P-S=\left(\tfrac12,-\tfrac12,-\tfrac12\right)p=P−S=(21​,−21​,−21​) q=Q−S=(−12,12,−12)q=Q-S=\left(-\tfrac12,\tfrac12,-\tfrac12\right)q=Q−S=(−21​,21​,−21​) r=R−S=(−12,−12,12)r=R-S=\left(-\tfrac12,-\tfrac12,\tfrac12\right)r=R−S=(−21​,−21​,21​) t=T−S=(12,12,12)t=T-S=\left(\tfrac12,\tfrac12,\tfrac12\right)t=T−S=(21​,21​,21​)

  1. Compute p×qp\times qp×q
\begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] \tfrac12 & -\tfrac12 & -\tfrac12\\[2pt] -\tfrac12 & \tfrac12 & -\tfrac12 \end{vmatrix}$$ $$=\hat i\left[\left(-\tfrac12\right)\left(-\tfrac12\right)-\left(-\tfrac12\right)\left(\tfrac12\right)\right] -\hat j\left[\left(\tfrac12\right)\left(-\tfrac12\right)-\left(-\tfrac12\right)\left(-\tfrac12\right)\right] +\hat k\left[\left(\tfrac12\right)\left(\tfrac12\right)-\left(-\tfrac12\right)\left(-\tfrac12\right)\right]$$ $$=\hat i\left(\tfrac14+\tfrac14\right)-\hat j\left(-\tfrac14-\tfrac14\right)+\hat k(\tfrac14-\tfrac14)$$ $$=\left(\tfrac12,\tfrac12,0\right)$$ 4. **Compute $r\times t$** $$r\times t= \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] -\tfrac12 & -\tfrac12 & \tfrac12\\[2pt] \tfrac12 & \tfrac12 & \tfrac12 \end{vmatrix}$$ $$=\hat i\left[\left(-\tfrac12\right)\left(\tfrac12\right)-\left(\tfrac12\right)\left(\tfrac12\right)\right] -\hat j\left[\left(-\tfrac12\right)\left(\tfrac12\right)-\left(\tfrac12\right)\left(\tfrac12\right)\right] +\hat k\left[\left(-\tfrac12\right)\left(\tfrac12\right)-\left(-\tfrac12\right)\left(\tfrac12\right)\right]$$ $$=\left(-\tfrac12,\tfrac12,0\right)$$ 5. **Compute $(p\times q)\times(r\times t)$** Let $$a=p\times q=\left(\tfrac12,\tfrac12,0\right),\qquad b=r\times t=\left(-\tfrac12,\tfrac12,0\right)$$ Then $$a\times b= \begin{vmatrix} \hat i & \hat j & \hat k\\[2pt] \tfrac12 & \tfrac12 & 0\\[2pt] -\tfrac12 & \tfrac12 & 0 \end{vmatrix}$$ $$=\hat i(0)-\hat j(0)+\hat k\left[\left(\tfrac12\right)\left(\tfrac12\right)-\left(\tfrac12\right)\left(-\tfrac12\right)\right]$$ $$=\hat k\left(\tfrac14+\tfrac14\right)=\left(0,0,\tfrac12\right)$$ 6. **Find the magnitude** $$\left| (p\times q)\times(r\times t) \right|=\left|\left(0,0,\tfrac12\right)\right|=\tfrac12$$ Hence the required integer-type value is $$\boxed{\tfrac12}$$
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