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: Set up the determinant with rows in the same order (a then b):
$\vec a\times\vec b$ with rows $[3,1,4]$ and $[1,-1,1]$ under $\hat i,\hat j,\hat k$.

Step 2: Expand along the $\hat j$ column carefully as a cross-check:
$\hat i$-component: $(1)(1)-(4)(-1)=1+4=5$. $\hat j$-component (with the minus sign): $-[(3)(1)-(4)(1)]=-[3-4]=1$. $\hat k$-component: $(3)(-1)-(1)(1)=-3-1=-4$. So $\vec a\times\vec b = 5\hat i+\hat j-4\hat k$, matching the standard expansion.

Step 3: Magnitude gives the area:
$|\vec a\times\vec b| = \sqrt{25+1+16}=\sqrt{42}$.

Final Answer:
Area $=\sqrt{42}$ square units. \[ \boxed{\sqrt{42} \text{ sq. units}} \]