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Question 112 of 162

Q.If PR⃗=2i⃗+j⃗+k⃗\vec{PR} = 2\vec i + \vec j + \vec k, QS⃗=−i⃗+3j⃗+2k⃗\vec{QS} = -\vec i + 3\vec j + 2\vec k then the area of the quadrilateral PQRS is :

(a) 535\sqrt3
(b) 10310\sqrt3
(c) 532\dfrac{5\sqrt3}{2}
(d) 32\dfrac{3}{2}
Puducherry TnboardTamil Nadu HSC (DGE) Board 2017MCQ· 1mImportance★★★★★
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The area of a quadrilateral in terms of its diagonal vectors d⃗1,d⃗2\vec d_1,\vec d_2 is 12∣d⃗1×d⃗2∣\frac12|\vec d_1\times\vec d_2|; here the diagonals are PR⃗\vec{PR} and QS⃗\vec{QS}, so compute their cross product and halve its magnitude.

  1. For quadrilateral PQRS, the diagonals are PRPR and QSQS, and the standard vector-algebra result is Area(PQRS)=12 ∣PR⃗×QS⃗∣.\text{Area}(PQRS) = \frac12\,|\vec{PR}\times\vec{QS}|.
  2. Given PR⃗=2i⃗+j⃗+k⃗\vec{PR}=2\vec i+\vec j+\vec k and QS⃗=−i⃗+3j⃗+2k⃗\vec{QS}=-\vec i+3\vec j+2\vec k.
  3. Compute the cross product: PR⃗×QS⃗=∣i⃗j⃗k⃗211−132∣\vec{PR}\times\vec{QS}=\begin{vmatrix}\vec i & \vec j & \vec k\\ 2 & 1 & 1\\ -1 & 3 & 2\end{vmatrix}
  4. i⃗\vec i-component: (1)(2)−(1)(3)=2−3=−1(1)(2)-(1)(3) = 2-3=-1.
  5. j⃗\vec j-component: −[(2)(2)−(1)(−1)]=−(4+1)=−5-[(2)(2)-(1)(-1)] = -(4+1)=-5. …

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