Find the area of the parallelogram whose adjacent sides are determined by the vector \(\vec{a}\) = \(\hat{i}\) - \(\hat{j}\) +3 \(\hat{k}\) and \(\vec{b}\) = 2\(\hat{i}\) - 7\(\hat{j}\) + \(\hat{k}\)

15√2 units

The  area of the parallelogram whose adjacent sides are the vector \(\vec{a}\) and \(\vec{b}\) is |\(\vec{a}\)x\(\vec{b}\)|

Adjacent sides are given as:

\(\vec{a}\) = \(\hat{i}\) - \(\hat{j}\) +3 \(\hat{k}\) and \(\vec{b}\) = 2\(\hat{i}\) - 7\(\hat{j}\) + \(\hat{k}\)

\(\vec{a}\)x \(\vec{b}\)  = (\(\hat{i}\) - \(\hat{j}\) +3 \(\hat{k}\)) x (2\(\hat{i}\) - 7\(\hat{j}\) + \(\hat{k}\))

\(\vec{a}\)x \(\vec{b}\) = (-1+21)\(\hat{i}\) - (1-6) \(\hat{j}\) + (-7+2) \(\hat{k}\)

\(\vec{a}\)x \(\vec{b}\) = 20\(\hat{i}\) +5 \(\hat{j}\) -5 \(\hat{k}\)

|\(\vec{a}\)x \(\vec{b}\)| = √(20)² +(5)² +(-5)² =15√2 

Hence the area of the parallelogram = 15√2 units