Question:
The area (in sq. units) of the parallelogram whose diagonals are along the vectors \(8\hat{i} - 6\hat{j}\) and \(3\hat{i} + 4\hat{j} - 12\hat{k}\), is :
The area (in sq. units) of the parallelogram whose diagonals are along the vectors \(8\hat{i} - 6\hat{j}\) and \(3\hat{i} + 4\hat{j} - 12\hat{k}\), is :
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Area = half the magnitude of the cross product of the diagonals: \(\frac12\sqrt{72^2+96^2+50^2} = 65\).
Updated On: Nov 11, 2026
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The Correct Option is B
Solution and Explanation
Area of a parallelogram with diagonals \(\vec d_1, \vec d_2\) is \(\frac12 |\vec d_1 \times \vec d_2|\).
With \(\vec d_1 = (8,-6,0)\), \(\vec d_2 = (3,4,-12)\): \(\vec d_1 \times \vec d_2 = \left((-6)(-12) - 0\cdot4,\ 0\cdot3 - 8(-12),\ 8\cdot4 - (-6)\cdot3\right) = (72, 96, 50)\).
Magnitude \(= \sqrt{5184 + 9216 + 2500} = \sqrt{16900} = 130\).
Area \(= \frac12 \times 130 = 65\).
Answer: (2) 65
With \(\vec d_1 = (8,-6,0)\), \(\vec d_2 = (3,4,-12)\): \(\vec d_1 \times \vec d_2 = \left((-6)(-12) - 0\cdot4,\ 0\cdot3 - 8(-12),\ 8\cdot4 - (-6)\cdot3\right) = (72, 96, 50)\).
Magnitude \(= \sqrt{5184 + 9216 + 2500} = \sqrt{16900} = 130\).
Area \(= \frac12 \times 130 = 65\).
Answer: (2) 65
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