Question

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

• A. $\frac{\sqrt{31}}{2}$
• B. $\frac{\sqrt{34}}{2}$
• C. $\frac{\sqrt{61}}{2}$
• D. $\frac{\sqrt{47}}{4}$

Answer 1

Answer: (C)

To determine the area of triangle, use formula $\Delta =\frac{1}{2}|AB\times AC|$.
The vertices of $△ABC$ are given as $A(1,1,2),B(2,3,5)$ and $C(1,5,5)$
First, we find vectors $AB$ and $AC$.
Now, $AB=PV\text{ of }B-PV\text{ of }A=(2\hat{i}+3\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k})$
$=(2-1)\hat{i}+(3-1)\hat{j}+(5-2)\hat{k}=\hat{i}+2\hat{j}+3\hat{k}$
and $AC=PV\text{ of }C-PV\text{ of }A$
$=(\hat{i}+5\hat{j}+5\hat{k})-(\hat{i}+\hat{j}+2\hat{k})$
$=(1-1)\hat{i}+(5-1)\hat{j}+(5-2)\hat{k}=4\hat{j}+3\hat{k}$
$\therefore AB\times AC=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ 0& 4& 3\end{array}\right|$
$=\hat{i}(6-12)-\hat{j}(3-0)+\hat{k}(4-0)=-6\hat{i}-3\hat{j}+4\hat{k}$
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 units.
Hence, the area of $△ABC$ is $\frac{\sqrt{61}}{2}$ sq units.