Question

Find the area of the parallelogram whose adjacent sides are given by \(\vec a = 3\hat i+\hat j+4\hat k\) and \(\vec b = \hat i-\hat j+\hat k\).

Hint:

Area of parallelogram = |a × b|; compute the cross product then its magnitude.

Answer 1

Step 1: Key Formula or Approach:
The area of a parallelogram with adjacent sides \(\vec a\) and \(\vec b\) is \(|\vec a \times \vec b|\).

Step 2: Computing the cross product:
\[ \vec a\times\vec b = \begin{vmatrix}\hat i & \hat j & \hat k\\3&1&4\\1&-1&1\end{vmatrix} = \hat i(1\cdot1-4\cdot(-1)) - \hat j(3\cdot1-4\cdot1) + \hat k(3\cdot(-1)-1\cdot1) \]
\[ = \hat i(1+4) - \hat j(3-4) + \hat k(-3-1) = 5\hat i + \hat j - 4\hat k \]

Step 3: Finding the magnitude:
\[ |\vec a\times\vec b| = \sqrt{5^2+1^2+(-4)^2} = \sqrt{25+1+16} = \sqrt{42} \]

Final Answer:
The area of the parallelogram is \(\sqrt{42}\) square units. \[ \boxed{\sqrt{42} \text{ sq. units}} \]