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Q.
The area of the triangle with vertices $A(1,1,2)$, $B(2,3,5)$ and $C(1,5,5)$ is
Vector Algebra
Solution:
To determine the area of triangle, use formula $\Delta=\frac{1}{2}| AB \times AC |$.
The vertices of $\triangle A B C$ are given as $A(1,1,2), B(2,3,5)$ and $C(1,5,5)$
First, we find vectors $A B$ and $A C$.
Now, $A B =P V \text { of } B-P V \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 $ A C =P V \text { of } C-P V \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 A B \times A C =\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\
1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} $
$ =\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 |A B \times A C|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6)^2+(-3)^2+(4)^2} $
$=\sqrt{36+9+16}=\sqrt{61}$
Area of $ \Delta A B C=\frac{1}{2}|A B \times A C|=\frac{1}{2} \times \sqrt{61}=\frac{\sqrt{61}}{2}$ sq units.
Hence, the area of $\triangle A B C$ is $\frac{\sqrt{61}}{2}$ sq units.