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Q9   Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

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Given in the question

vertices A=(1, 1, 2), B=(2, 3, 5) and C=(1, 5, 5). and we need to find the area of the triangle

AB=(2-1)\hat i+(3-1)\hat j+(5-2)\hat k=\hat i+2\hat j+3\hat k

BC=(1-2)\hat i+(5-3)\hat j+(5-5)\hat k=-\hat i+2\hat j

Now as we know

Area of triangle 

A=\frac{1}{2}|\vec {AB}\times\vec {BC}|=\frac{1}{2}|(\hat i+2\hat j +3\hat k)\times(-\hat i+2\hat j)|

\\A=\frac{1}{2}\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &2 &3 \\ -1 &2 &0 \end{vmatrix}=\frac{1}{2}|\hat i(0-6)-\hat j(0-(-3))+\hat k(2-(-2))| \\A=\frac{1}{2}|-6\hat i-3\hat j+4\hat k|


 The area of the triangle is \frac{\sqrt{61}}{2} square units

Posted by

Pankaj Sanodiya

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