Question

Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5)

Answer 1

The vertices of triangle ABC are given as A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
First we find vectors AB and AC.
Now, AB = PV of B - PV of A $=(2\hat{i}+3\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k})=(2-1)\hat{i}+(3-1)\hat{j}+(5-2)\hat{k}=\hat{i}+2\hat{j}+3\hat{k}$
and AC = PV of C - PV of A
$=(\hat{i}+5\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k})=(1-1)\hat{i}+(5-1)\hat{j}+(5-2)\hat{k}=4\hat{j}+3\hat{k}\therefore AB\times AC=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 0& 4& 3\end{array}\right|=\hat{i}(6-12)-\hat{j}(3-0)+\hat{k}(4-0)=-6\hat{i}-3\hat{j}+4\hat{k}$
Comparing with $X=x\hat{i}+y\hat{j}+z\hat{k}$, we get $x=-6,y=-3,z=4$
$\therefore |AB\times AC|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6{)}^2+(-3{)}^2+(4{)}^2}=\sqrt{36+9+16}=\sqrt{61}$
Area of $\Delta ABC=\frac{1}{2}|AB\times AC|=\frac{1}{2}\times \sqrt{61}=\frac{\sqrt{61}}{2}sq.unit$
Hence, the area of $\Delta ABC$ is $\frac{\sqrt{61}}{2}sq.unit$.