Chapter 11, Problem 55RE

a.

To determine

To find: If the given points are vertices of a parallelogram.

Answer to Problem 55RE

  $ABDC$ is a parallelogram.

Explanation of Solution

Given information:

Four points are given as $\left(2,-1,1\right),\left(5,1,4\right),\left(0,1,1\right),\left(3,3,4\right)$ .

Concepts used:

Area of a parallelogram is given as − $\Vert \overrightarrow{AB}\times \overrightarrow{AD}\Vert$ where $\overrightarrow{AB}\text{ and }\overrightarrow{AD}$ are adjacent sides.

Calculation:

Let us suppose, $A\left(2,-1,1\right),B\left(5,1,4\right),C\left(0,1,1\right),D\left(3,3,4\right)$

  $\begin{array}{l}\text{Now, }\overrightarrow{AB}=\left(5-2\right)i+\left(1+1\right)j+\left(4-1\right)k\\ \Rightarrow \overrightarrow{AB}=3i+2j+3k\\ \overrightarrow{DC}=\left(0-3\right)i+\left(1-3\right)j+\left(1-4\right)k\\ \Rightarrow \overrightarrow{DC}=-3i-2j-3k\\ \Rightarrow \overrightarrow{AB}=-\overrightarrow{DC}\text{ These two are parallel and opposite direction}\text{.}\end{array}$

  $\begin{array}{l}\text{Similarly, }\overrightarrow{AC}=\left(0-2\right)i+\left(1+1\right)j+\left(1-1\right)k\\ \Rightarrow \overrightarrow{AC}=-2i+2j\\ \overrightarrow{DB}=\left(5-3\right)i+\left(1-3\right)j+\left(4-4\right)k\\ \Rightarrow \overrightarrow{DB}=2i-2j\\ \Rightarrow \overrightarrow{AC}=-\overrightarrow{DB}\text{ These two are parallel and opposite direction}\text{.}\end{array}$

Hence, $ABDC$ is a parallelogram with $AB$ and $AC$ as adjacent sides.

b.

To determine

To find: Area of the parallelogram.

Answer to Problem 55RE

  $2\sqrt{43}\text{ square unit}\text{. }$

Explanation of Solution

Given information:

Four points are given as $\left(2,-1,1\right),\left(5,1,4\right),\left(0,1,1\right),\left(3,3,4\right)$ .

Concepts used:

Area of a parallelogram is given as − $\Vert \overrightarrow{AB}\times \overrightarrow{AD}\Vert$ where $\overrightarrow{AB}\text{ and }\overrightarrow{AD}$ are adjacent sides.

Calculation:

Let us suppose, $A\left(2,-1,1\right),B\left(5,1,4\right),C\left(0,1,1\right),D\left(3,3,4\right)$

From part $a$ of this exercise, $ABDC$ is a parallelogram with $AB$ and $AC$ as adjacent sides.

  $\overrightarrow{AB}=3i+2j+3k$ and $\overrightarrow{AC}=-2i+2j$

  $\begin{array}{l}\text{Now, }\overrightarrow{AB}\times \overrightarrow{AC}=\left|\begin{array}{ccc}i& j& k\\ 3& 2& 3\\ -2& 2& 0\end{array}\right|\\ \Rightarrow \overrightarrow{AB}\times \overrightarrow{AC}=i\left(0-6\right)-j\left(0+6\right)+k\left(6+4\right)\\ \Rightarrow \overrightarrow{AB}\times \overrightarrow{AC}=-6i-6j+10k\end{array}$

Area of parallelogram will be −

  $\begin{array}{l}\Vert \overrightarrow{AB}\times \overrightarrow{AC}\Vert =\sqrt{{\left(-6\right)}^2+{\left(-6\right)}^2+{10}^2}\\ \Rightarrow \Vert \overrightarrow{AB}\times \overrightarrow{AC}\Vert =2\sqrt{43}\text{ square unit}\text{. }\end{array}$

Hence, area of parallelogram $ABDC$ is $2\sqrt{43}\text{ square unit}\text{. }$