The diagonals of a parallelograms are given by the vectors $3\vec {i} + \vec {j} + 2\vec {k}$ and $\vec {i} – 3\vec {j} + 4\vec {k}$. Find the area of the parallelogram.
My Attempt:
Let $\vec {d_1}=3\vec {i} + \vec {j} + 2\vec {k}$ and $\vec {d_2}=\vec {i} – 3\vec {j} + 4\vec {k}$ be two diagonals represented in vector form.
How do I get the base and altitude to find the area of parallelogram?
Best Answer
Let $\vec a=3\vec {i} + \vec {j} + 2\vec {k}$
$\vec b=\vec {i} - 3\vec {j} + 4\vec {k}$ be the diagonals of a parallelogram, then its vector area $=\dfrac12(\vec a\times \vec b)$
Here $\vec a\times \vec b=\begin{vmatrix} \vec {i} & \vec {j} & \vec {k} \\ 3 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix}$ $$=\vec {i}(4-6)-\vec {j}(12+2)+\vec {k}(-9-1)$$ $$=-2\vec {i}-14\vec {j}-10\vec {k}$$ $$=2(-\vec {i}-7\vec {j}-5\vec {k})$$ Therefor, vector area of the parallelogram is $$=\frac12[2(-\vec {i}-7\vec {j}-5\vec {k})]$$ $$-\vec {i}-7\vec {j}-5\vec {k}$$ Area of the parallelogram is $$=|-\vec {i}-7\vec {j}-5\vec {k}|$$ $$=\sqrt{1+49+25}=\sqrt{75}=5\sqrt{3}\mbox{ sq. units}$$