Chapter 11.3, Problem 45E

To determine

ToEvaluate:The area of the triangle with vertices $\left(0,0,0\right)$ , $\left(1,2,3\right)$ and $\left(3,0,0\right)$

Answer to Problem 45E

The area of the triangle is $A=\frac{1}{2}\sqrt{117}\text{ sq units}$

Explanation of Solution

Given:

The vertices of triangle $\left(0,0,0\right)$ , $\left(1,2,3\right)$ and $\left(3,0,0\right)$

Concept Used:

For the two adjacent sides of a triangle $u=\langle u_1,u_2,u_3\rangle$ and $v=\langle v_1,v_2,v_3\rangle$ , the areacan be calculated as $A=\frac{1}{2}\Vert u\times v\Vert$ where $u\times v=\left|\begin{array}{ccc}i& j& k\\ u_1& u_2& u_3\\ v_1& v_2& v_3\end{array}\right|$

Calculation:

Forthe vertices of triangle $\left(0,0,0\right)$ , $\left(1,2,3\right)$ and $\left(3,0,0\right)$

Let $A=\left(0,0,0\right)$ , $B=\left(1,2,3\right)$ and $C=\left(3,0,0\right)$

Then two adjacent sides of triangle are:

  $\begin{array}{c}\overrightarrow{AB}=\langle 1-0,2-0,3-0\rangle \\ =\langle 1,2,3\rangle \end{array}$

and

  $\begin{array}{c}\overrightarrow{AC}=\langle 3-0,0-0,0-0\rangle \\ =\langle 3,0,0\rangle \end{array}$

  $\begin{array}{c}\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}i& j& k\\ 1& 2& 3\\ 3& 0& 0\end{array}\right|\\ =i\left|\begin{array}{cc}2& 3\\ 0& 0\end{array}\right|-j\left|\begin{array}{cc}1& 3\\ 3& 0\end{array}\right|+k\left|\begin{array}{cc}1& 2\\ 3& 0\end{array}\right|\\ =i\left(\left(2\right)\left(0\right)-\left(0\right)\left(3\right)\right)-j\left(\left(1\right)\left(0\right)-\left(3\right)\left(3\right)\right)+k\left(\left(1\right)\left(0\right)-\left(3\right)\left(2\right)\right)\\ =i\left(0-0\right)-j\left(0-9\right)+k\left(0-6\right)\end{array}$

  $=0i+9j-6k$

The area is $A=\frac{1}{2}\Vert \overrightarrow{AB}\times \overrightarrow{AC}\Vert$

  $\begin{array}{c}A=\frac{1}{2}\sqrt{{\left(0\right)}^2+{\left(9\right)}^2+{\left(-6\right)}^2}\\ =\frac{1}{2}\sqrt{0+81+36}\\ =\frac{1}{2}\sqrt{117}\end{array}$

Therefore, the area of the triangle is $A=\frac{1}{2}\sqrt{117}\text{ sq units}$

Conclusion:

The area of the triangle is $A=\frac{1}{2}\sqrt{117}\text{ sq units}$