Given two diagonals of a parallelogram are \(2\hat i+3\hat j+6\hat k\) and \(3\hat i-6\hat j+2\hat k\)
Let \(\vec a=2\hat i+3\hat j+6\hat k\) and \(\vec b=3\hat i-6\hat j+2\hat k\)
Recall the area of the parallelogram whose diagonals are given by the two vectors \(\vec a=a_1\hat i+a_2\hat j+a_3\hat k\) and \(\vec b = b_1\hat i+b_2\hat j+b_3\hat k\) is \(\cfrac12|\vec a\times\vec b|\) where
Here, we have (a1, a2, a3) = (2, 3, 6) and (b1, b2, b3) = (3, –6, 2)
Recall the magnitude of the vector \(\text x\hat i+y\hat j+z\hat k\) is
Thus, area of the parallelogram is 24.5 square units.