Chapter 9, Problem 108RE

To determine

To calculate: The area of the parallelogram with vertices,

  $P_1=\left(2,-1,1\right),\text{}{P}_2=\left(5,1,4\right),P_3=\left(0,1,1\right)\text{}and{P}_4=\left(3,3,4\right)$

Answer to Problem 108RE

The area pf parallelogram is $\text{14}\text{square}\text{unit}$ .

Explanation of Solution

Given information: The vertices are given as:

  $P_1=\left(2,-1,1\right),\text{}{P}_2=\left(5,1,4\right),P_3=\left(0,1,1\right)\text{}and{P}_4=\left(3,3,4\right)$

Formula used:

let $u\text{and}\text{}\text{}v$ are adjacent sides of parallelogram, area of parallelogram $=\Vert u\Vert \left[\Vert v\Vert \sin \theta \right]=\Vert u\times v\Vert$

where $\Vert u\Vert \text{and}\text{}\text{}\Vert v\Vert$ are lengths of sides, $\theta$ is the angle between $u\text{and}\text{}\text{}v$ ,height of the parallelogram is $\Vert v\Vert \sin \theta$ .

The position vector $v=\overrightarrow{P_1{P}_2}=\left(x_2-x_1\right)i+\left(y_2-y_1\right)j+\left(z_2-z_1\right)k$

Where $P_1=\left(x_1,y_1,z_1\right)$ is initial point and $P_2=\left(x_2,y_2,z_2\right)$ is terminal point.

The cross product of vectors $u=a_{1}i+b_{1}j+c_{1}k$ and $v=a_{2}i+b_{2}j+c_{2}k$ is,

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|\\ =i\left|\begin{array}{cc}{b}_1& c_1\\ b_2& c_2\end{array}\right|-j\left|\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right|+k\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|\\ =i\left(b_1{c}_2-c_1{b}_2\right)-j\left(a_1{c}_2-c_1{a}_2\right)+k\left(a_1{b}_2-b_1{a}_2\right)\end{array}$

Calculation:

Consider the vertices,

  $P_1=\left(2,-1,1\right),\text{}{P}_2=\left(5,1,4\right),P_3=\left(0,1,1\right)\text{}and{P}_4=\left(3,3,4\right)$

The adjacent sides of the parallelogram are,

  $\begin{array}{c}u=\overrightarrow{P_1{P}_2}=\left(5-1\right)i+\left(1+1\right)j+\left(4-1\right)k\\ =4i+2j+3k\end{array}$

And

  $\begin{array}{c}v=\overrightarrow{P_1{P}_3}=\left(0-2\right)i+\left(1+1\right)j+\left(1-1\right)k\\ =-2i+2j+0k\end{array}$

Since, the cross product of vectors $u=4i+2j+3k$ and $v=-2i+2j+0k$ is,

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ 4& 2& 3\\ -2& 2& 0\end{array}\right|\\ =i\left|\begin{array}{cc}2& 3\\ 2& 0\end{array}\right|-j\left|\begin{array}{cc}4& 3\\ -2& 0\end{array}\right|+k\left|\begin{array}{cc}4& 2\\ -2& 2\end{array}\right|\\ =i\left(2-6\right)-j\left(0+6\right)+k\left(8+4\right)\\ =i\left(-4\right)-j\left(6\right)+k\left(12\right)\end{array}$

Further, the area of parallelogram

  $\begin{array}{l}\Vert u\times v\Vert =\sqrt{{\left(-4\right)}^2+{\left(-6\right)}^2+{\left(12\right)}^2}\\ =\sqrt{16+36+144}\\ =\sqrt{196}\\ =14\text{square}\text{unit}\end{array}$

Thus, the area of parallelogram is $\text{14}\text{}\text{square}\text{unit}$ .