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Find the area of triangle having the points as A(1,1,1), B(1,2,3) and C(2,3,1) as its vertices.

1 Answer

  • Area of the given triangle is $\large\frac{1}{2}$$\mid \overrightarrow {AB}\times \overrightarrow{AC}\mid$
Step 1:
Let $A(1,1,1),B(1,2,3)$ and $C(2,3,1)$
$\quad\;\;=(\hat i+2\hat j+3\hat k)-(\hat i+\hat j+\hat k)$
$\quad\;\;=\hat j+2\hat k$
$\quad\;\;=(2\hat i+3\hat j-\hat k)-(\hat i+\hat j+\hat k)$
$\quad\;\;=\hat i+2\hat j$
Step 2:
Area of the given triangle is $\large\frac{1}{2}$$\mid \overrightarrow {AB}\times \overrightarrow{AC}\mid$
$\overrightarrow {AB}\times \overrightarrow{AC}=\begin{vmatrix}\hat i&\hat j&\hat k\\0&1&2\\1&2&0\end{vmatrix}$
$\qquad\qquad=-4\hat i+2\hat j-\hat k$
$\therefore \mid \overrightarrow {AB}\times \overrightarrow{AC}\mid=\sqrt{(-4)^2+(2)^2+(-1)^2}$
Hence the required area is $\large\frac{1}{2}$$\sqrt{21}$
answered Nov 8, 2013 by sreemathi.v

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