Chapter 11.3, Problem 50E

To determine

The area of the triangle with the vertices $\left(1,-4,3\right),\left(2,0,2\right),\left(-2,2,0\right)$ .

Answer to Problem 50E

The area of triangleis $3\sqrt{11}$ .

Explanation of Solution

Given information:

The three vertices of the triangle $\left(1,-4,3\right),\left(2,0,2\right),\left(-2,2,0\right)$ .

Formula used:

If $a\left(a_1,a_2,a_3\right)\text{, }b\left(b_1,b_2,b_3\right)\text{ and}c\left(c_1,c_2,c_3\right)$ are the three vertices of the triangle, then the vectors that represents the sides of the triangle are calculated as,

  $\begin{array}{l}\overrightarrow{AB}=\left(b_1-a_1\right)i+\left(b_2-a_2\right)j+\left(b_3-a_3\right)k\\ \overrightarrow{AC}=\left(c_1-a_1\right)i+\left(c_2-a_2\right)j+\left(c_3-a_3\right)k\end{array}$

If $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are sides of the triangle then the area of triangle with its sides as $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is equal to the half of their magnitude of their cross product $\frac{1}{2}\Vert \overrightarrow{AB}\times \overrightarrow{AC}\Vert$ .

Let $\overrightarrow{AB}\times \overrightarrow{AC}$ represents the vector $x$ such that $x=x_{1}i+x_{2}j+x_{3}k$ then the magnitude of the vector $x$ is given by $\Vert x\Vert =\sqrt{x_1{}^2+x_2{}^2+x_3{}^3}$ .

If there are two three-dimensional vectors in a space, say, $b=b_{1}i+b_{2}j+b_{3}k$ and $a=a_{1}i+a_{2}j+a_{3}k$ , then the cross product of two vectors is defined as ,

  $\begin{array}{c}a\times b=\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\\ \text{= }\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k\end{array}$

Calculation:

Consider the given vertices of the triangle $\left(1,-4,3\right),\left(2,0,2\right),\left(-2,2,0\right)$ .

Let the given vertices $A\left(1,-4,3\right),B\left(2,0,2\right),C\left(-2,2,0\right)$ .

Recall that if $a\left(a_1,a_2,a_3\right)\text{, }b\left(b_1,b_2,b_3\right)\text{ and}c\left(c_1,c_2,c_3\right)$ are the three vertices of the triangle, then the vectors that represents the sides of the triangle are calculated as,

  $\begin{array}{l}\overrightarrow{AB}=\left(b_1-a_1\right)i+\left(b_2-a_2\right)j+\left(b_3-a_3\right)k\\ \overrightarrow{AC}=\left(c_1-a_1\right)i+\left(c_2-a_2\right)j+\left(c_3-a_3\right)k\end{array}$

So, the vertices of the triangle will be calculated as,

  $\begin{array}{c}\overrightarrow{AB}=\left(b_1-a_1\right)i+\left(b_2-a_2\right)j+\left(b_3-a_3\right)k\\ =\left(2-1\right)i+\left(0-\left(-4\right)\right)j+\left(2-3\right)k\\ =i+4j-k\end{array}$

  $\begin{array}{c}\overrightarrow{AC}=\left(c_1-a_1\right)i+\left(c_2-a_2\right)j+\left(c_3-a_3\right)k\\ =\left(-2-1\right)i+\left(2-\left(-4\right)\right)j+\left(0-3\right)k\\ =-3i+6j-3k\end{array}$

Recall if there are two three-dimensional vectors in a space, say, $b=b_{1}i+b_{2}j+b_{3}k$ and $a=a_{1}i+a_{2}j+a_{3}k$ , then the cross product of two vectors is defined as ,

  $\begin{array}{c}a\times b=\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\\ \text{= }\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k\end{array}$

So, the cross product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is calculated as,

  $\begin{array}{c}\overrightarrow{AB}\times \overrightarrow{AC}=\begin{array}{ccc}i& j& k\\ 1& 4& -1\\ -3& 6& -3\end{array}\\ \text{= }\left(\left[4\times \left(-3\right)\right]-\left[6\times \left(-1\right)\right]\right)i-\left(\left[\left(-3\right)\times 1\right]-\left[\left(-3\right)\times \left(-1\right)\right]\right)j+\left(6\times 1-\left[4\times \left(-3\right)\right]\right)k\\ =\left[-12-\left(-6\right)\right]i-\left[-3-3\right]j+\left[6-\left(-12\right)\right]k\\ =-6i+6j+18k\end{array}$

Recall that if $\overrightarrow{AB}\times \overrightarrow{AC}$ represents the vector $x$ such that $x=x_{1}i+x_{2}j+x_{3}k$ then the magnitude of the vector $x$ is given by $\Vert x\Vert =\sqrt{x_1{}^2+x_2{}^2+x_3{}^3}$ .

So, the magnitude of $\overrightarrow{AB}\times \overrightarrow{AC}$ will be calculated as,

  $\begin{array}{c}\Vert \overrightarrow{AB}\times \overrightarrow{AC}\Vert =\sqrt{{\left(-6\right)}^2+6^2+{18}^2}\\ =\sqrt{36+36+324}\\ =\sqrt{396}\\ =6\sqrt{11}\end{array}$

Now, recall if $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are sides of the triangle then the area of triangle with its sides as $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is equal to the half of their magnitude of their cross product $\frac{1}{2}\Vert \overrightarrow{AB}\times \overrightarrow{AC}\Vert$ .

So, the area of the triangle will be calculated as,

  $\begin{array}{c}ar\left(\Delta \right)=\frac{1}{2}\Vert \overrightarrow{AB}\times \overrightarrow{AC}\Vert \\ =\frac{1}{2}\times 6\sqrt{11}\\ =3\sqrt{11}\end{array}$

Hence, the area of triangleis $3\sqrt{11}$ .