Given, two adjacent sides of the parallelogram
\(2\hat i - 4\hat j + 5\hat k \) and \(\hat i - 2\hat j - 3\hat k\)
The diagonal will be the resultant of these two vectors.
So resultant R:
\(\vec R = 2\hat i - 4\hat j + 5\hat k + \hat i -2 \hat j - 3 \hat k\)
\(= 3\hat i - 6\hat j +2 \hat k\)
Now unit vector in direction of R
\(\vec u = \frac{3\hat i - 6\hat j + 2\hat k}{\sqrt{3^2 + (-6)^2 + 2^2}}\)
\(= \frac{3\hat i - 6\hat j + 2 \hat k}{\sqrt{49}}\)
\(= \frac{3\hat i - 6\hat j + 2\hat k}{7}\)
Hence unit vector along the diagonal of the parallelogram
\(\vec u= \frac 37 \hat i - \frac 67 \hat j + \frac 27 \hat k\)
Now,
Area of parallelogram
\(A = (2\hat i - 4\hat j + 5\hat k) \times (\hat i - 2 \hat j - 3 \hat k)\)
\(A = \begin{vmatrix}\hat i &\hat j & \hat k\\2&-4&5\\1&-2&-3\end{vmatrix}\)
\(= |\hat i (12 + 10) - \hat j (-6 - 5) + \hat k(-4 + 4)|\)
\(= |22\hat i + 11\hat j|\)
\(A = \sqrt{22^2 + 11^2 }\)
\(= 11\sqrt 5\)
Hence the area of the parallelogram is 11√5.