Chapter 9.5, Problem 26E

To determine

To find: The area of triangle formed by the given vectors.

Answer to Problem 26E

The area of triangle formed by the given vectors is $\frac{\sqrt{101}}{2}$ .

Explanation of Solution

Given information:

The points are $P\left(2,1,0\right),Q\left(0,0,-1\right),R\left(-4,2,0\right)$ .

Calculation:

The length of the vector $\overrightarrow{PQ}$ is,

  $\begin{array}{c}\overrightarrow{PQ}=\left(0-2\right)i+\left(0-1\right)j+\left(-1-0\right)k\\ =-2i-j-k\end{array}$

The length of the vector $\overrightarrow{PR}$ is,

  $\begin{array}{c}\overrightarrow{PR}=\left(-4-2\right)i+\left(2-1\right)j+\left(0-0\right)k\\ =-6i+j\end{array}$

The cross product is given as,

  $a\times b=\left|\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|$

Substitute the values in above equation.

  $\begin{array}{c}\overrightarrow{PQ}\times \overrightarrow{PR}=\left|\begin{array}{ccc}i& j& k\\ -2& -1& -1\\ -6& 1& 0\end{array}\right|\\ =i+6j-8k\end{array}$

The half of magnitude of cross product gives the area of triangle.

  $\begin{array}{c}\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|=\frac{1}{2}\sqrt{{\left(1\right)}^2+{\left(6\right)}^2+{\left(-8\right)}^2}\\ =\frac{\sqrt{101}}{2}\end{array}$