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If a⃗=2i^−3j^+k^,b⃗=−i^+k^,c⃗=2j^−k^\vec a=2\hat i-3\hat j+\hat k, \vec b=-\hat i+\hat k, \vec c=2\hat j-\hat k are three vectors, then area of the parallelogram having diagonals (a⃗+b⃗)(\vec a+\vec b) and (b⃗+c⃗)(\vec b+\vec c) is

Solution

✅ Correct Option: 3

a⃗+b⃗=i^−3j^+2k^\vec a+\vec b=\hat i-3\hat j+2\hat k and b⃗+c⃗=−i^+2j^\vec b+\vec c=-\hat i+2\hat j.

Their cross product is −4i^−2j^−k^-4\hat i-2\hat j-\hat k, of magnitude 21\sqrt{21}.

When the diagonals are given, the area is half this, namely 1221\frac{1}{2}\sqrt{21} square units.

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