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

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

Answers (1)

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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|

A=\frac{1}{2}*\sqrt{(-6)^2+(-3)^2+(4)^2}=\frac{\sqrt{61}}{2}

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

Posted by

Pankaj Sanodiya

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