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If \(\vec A=\hat i+2\hat j+3\hat k\) and \(\vec B=3\hat i-2\hat j+\hat k\), then the area of the parallelogram formed by \(\vec A\) and \(\vec B\) is:

1. 3
2. \(8\sqrt{3}\)
3. 64
4. 0

Correct answer: \(8\sqrt{3}\)

Solution

The area of a parallelogram formed by two vectors is \(|\vec A\times\vec B|\). For the given vectors, the cross product has magnitude \(8\sqrt{3}\), so that is the required area.

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