Question

Find the area of the triangle whose two adjecent sides are represented by the vectors $\overline{a}=3\overline{i}+4\overline{j}$ and $\overline{b}=-5\overline{i}+7\overline{j}$

Answer 1

As we know for a triangle formed by two vectors

Area of $△=\frac{|\overline{a}\times \overline{b}|}{2}$ .....(1)

We know that,

$\overline{a}\times \overline{b}=\left|\begin{array}{ccc}\overline{i}& \overline{j}& \overline{k}\\ 3& 4& 0\\ -5& 7& 0\end{array}\right|$

$\overline{a}\times \overline{b}=(4\times 0-7\times 0)\overline{i}+[3\times 0-(-5)\times 0]\overline{j}+[3\times 7-4\times (-5\left)\right]\overline{k}$

$\overline{a}\times \overline{b}=[21+20]\overline{k}$

$\overline{a}\times \overline{b}=41\overline{k}$

Now,

$|\overline{a}\times \overline{b}|=\sqrt{0^2+0^2+{41}^2}=41$

Substituting this value in equation(1), we get

Area of $△=\frac{41}{2}$

Hence, area of $△=\frac{41}{2}uni{t}^2$