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Q.The diagonals of a parallelogram are represented by the vectors d₁=2î-ĵ+k̂ and d₂=3î+4ĵ-k̂, then the area of the parallelogram is -

(a) √155 sq. units
(b) (1/2)√155 sq. units
(c) 2√155 sq. units
(d) (1/4)√155 sq. units
Mizoram MbseMizoram Board of School Education HSSLC 2024MCQ· 1mImportance★★★★★
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The area of a parallelogram from its diagonals d1⃗,d2⃗\vec{d_1},\vec{d_2} is 12∣d1⃗×d2⃗∣\frac{1}{2}|\vec{d_1}\times\vec{d_2}|.

Given d1⃗=2i^−j^+k^\vec{d_1} = 2\hat{i}-\hat{j}+\hat{k} and d2⃗=3i^+4j^−k^\vec{d_2} = 3\hat{i}+4\hat{j}-\hat{k}.

d1⃗×d2⃗=i^[(−1)(−1)−(1)(4)]−j^[(2)(−1)−(1)(3)]+k^[(2)(4)−(−1)(3)]\vec{d_1}\times\vec{d_2} = \hat{i}[(-1)(-1)-(1)(4)] - \hat{j}[(2)(-1)-(1)(3)] + \hat{k}[(2)(4)-(-1)(3)]

=i^(1−4)−j^(−2−3)+k^(8+3)=−3i^+5j^+11k^= \hat{i}(1-4) - \hat{j}(-2-3) + \hat{k}(8+3) = -3\hat{i}+5\hat{j}+11\hat{k}

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