Question

If $2\hat{i}+3\hat{j}+4\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ are two adjacent sides of a parallelogram, then the area of the parallelogram will be

• A. $\sqrt{87}$
• B. $\sqrt{70}$
• C. Cannot be calculated from the given data
• D. none of the above

Answer 1

The correct option is B $\sqrt{70}$

We know that when two adjacent sides of a parallelogram are given, the area of the parallelogram is given by the magnitude of the cross product of these two vectors. Here two given sides are $2\hat{i}+3\hat{j}+4\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$
So their cross product will be given by the following determinant
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 1& -1& 1\end{array}\right|$
$=\hat{i}\left(\right(3\times 1)-(4\times -1)-\hat{j}(2\times 1)-(4\times 1)+\hat{k}(2\times -1)-(3\times 1\left)\right)$
=$7\hat{i}+2\hat{j}-5\hat{k}$
Now the modulus of this cross product i.e. $7\hat{i}+2\hat{j}-5\hat{k}$, will give you the area of the parallelogram.
$|7\hat{i}+2\hat{j}-5\hat{k}|=\sqrt{(7{)}^2+(2{)}^2+(-5{)}^2}$
=$\sqrt{49+4+25}$
=$\sqrt{78}$
Which is nothing but the area of the given parallelogram