Chapter 11.3, Problem 42E

$\left(a\right)$

To determine

To verify: The points $A\left(1,1,1\right),B\left(6,5,2\right),C\left(7,7,5\right),D\left(2,3,4\right)$ are the vertices of the parallelogram.

Explanation of Solution

Given information:

The points $A\left(1,1,1\right),B\left(6,5,2\right),C\left(7,7,5\right),D\left(2,3,4\right)$ .

Formula used:

Four vertices A, B, C, and D are said to be vertices of parallelogram ABCD if $\stackrel{\rightharpoonup }{CD}=-\stackrel{\rightharpoonup }{AB}$ and $\stackrel{\rightharpoonup }{CB}=-\stackrel{\rightharpoonup }{AD}$ that is $\stackrel{\rightharpoonup }{AB}$ parallel to $\stackrel{\rightharpoonup }{CD}$ and $\stackrel{\rightharpoonup }{AD}$ is parallel to $\stackrel{\rightharpoonup }{CB}$ . Therefore, $\stackrel{\rightharpoonup }{AB}$ and $\stackrel{\rightharpoonup }{AD}$ are adjacent sides of parallelogram.

Proof:

Consider the points $A\left(1,1,1\right),B\left(6,5,2\right),C\left(7,7,5\right),D\left(2,3,4\right)$ .

Recall that four vertices A, B, C, and D are said to be vertices of parallelogram ABCD if $\stackrel{\rightharpoonup }{CD}=-\stackrel{\rightharpoonup }{AB}$ and $\stackrel{\rightharpoonup }{CB}=-\stackrel{\rightharpoonup }{AD}$ that is $\stackrel{\rightharpoonup }{AB}$ parallel to $\stackrel{\rightharpoonup }{CD}$ and $\stackrel{\rightharpoonup }{AD}$ is parallel to $\stackrel{\rightharpoonup }{CB}$ . Therefore, $\stackrel{\rightharpoonup }{AB}$ and $\stackrel{\rightharpoonup }{AD}$ are adjacent sides of parallelogram.

Now, the value of $\stackrel{\rightharpoonup }{AD}$ is evaluated as, subtract the coordinate of D from A and write it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AD}=\left(2-1\right)i+\left(3-1\right)j+\left(4-1\right)k\\ =i+2j+3k\end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{DB}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{DB}=\left(6-2\right)i+\left(5-3\right)j+\left(2-4\right)k\\ =4i+2j-2k\end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{BC}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{BC}=\left(7-6\right)i+\left(7-5\right)j+\left(5-2\right)k\\ =i+2j+3k\end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{CA}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{CA}=\left(7-1\right)i+\left(7-1\right)j+\left(5-1\right)k\\ =6i+6j+4k\end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{AB}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AB}=\left(6-1\right)i+\left(5-1\right)j+\left(2-1\right)k\\ =5i+4j+k\end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{DC}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{DC}=\left(7-2\right)i+\left(7-3\right)j+\left(5-4\right)k\\ =5i+4j+k\end{array}$

It is observed that,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AD}=\stackrel{\rightharpoonup }{BC}\\ \stackrel{\rightharpoonup }{AB}=\stackrel{\rightharpoonup }{DC}\end{array}$

It can be said that, $\stackrel{\rightharpoonup }{AD}$ is parallel to $\stackrel{\rightharpoonup }{BC}$ and $\stackrel{\rightharpoonup }{AB}$ is parallel to $\stackrel{\rightharpoonup }{DC}$ .

Thus, the quadrilateral formed by the points $A\left(1,1,1\right),B\left(6,5,2\right),C\left(7,7,5\right),D\left(2,3,4\right)$ represent a parallelogram.

Hence, verified.

$\left(b\right)$

To determine

To calculate:The area of the parallelogram.

Answer to Problem 42E

The area of the parallelogramis $2\sqrt{83}$ .

Explanation of Solution

Given information:

The points $A\left(1,1,1\right),B\left(2,3,4\right),C\left(6,5,2\right),D\left(7,7,5\right)$ .

Formula used:

Let $a\times b$ 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 given in the form of their components, say, $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ , then the cross product of two vectors is defined as ,

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

If $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ are the adjacent sides of a parallelogram then area of the parallelogram is $\Vert a\times b\Vert$ .

Calculation:

Consider the points $A\left(1,1,1\right),B\left(6,5,2\right),C\left(7,7,5\right),D\left(2,3,4\right)$ .

Recall that four vertices A, B, C, and D are said to be vertices of parallelogram ABCD if $\stackrel{\rightharpoonup }{CD}=-\stackrel{\rightharpoonup }{AB}$ and $\stackrel{\rightharpoonup }{CB}=-\stackrel{\rightharpoonup }{AD}$ that is $\stackrel{\rightharpoonup }{AB}$ parallel to $\stackrel{\rightharpoonup }{CD}$ and $\stackrel{\rightharpoonup }{AD}$ is parallel to $\stackrel{\rightharpoonup }{CB}$ . Therefore, $\stackrel{\rightharpoonup }{AB}$ and $\stackrel{\rightharpoonup }{AD}$ are adjacent sides of parallelogram.

Now, the value of $\stackrel{\rightharpoonup }{AD}$ is evaluated as, subtract the coordinate of D from A and write it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AD}=\left(2-1\right)i+\left(3-1\right)j+\left(4-1\right)k\\ =i+2j+3k\end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{DB}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{DB}=\left(6-2\right)i+\left(5-3\right)j+\left(2-4\right)k\\ =4i+2j-2k\end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{BC}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{BC}=\left(7-6\right)i+\left(7-5\right)j+\left(5-2\right)k\\ =i+2j+3k\end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{CA}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{CA}=\left(7-1\right)i+\left(7-1\right)j+\left(5-1\right)k\\ =6i+6j+4k\end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{AB}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AB}=\left(6-1\right)i+\left(5-1\right)j+\left(2-1\right)k\\ =5i+4j+k\end{array}$

Similarly, the vector $\stackrel{\rightharpoonup }{DC}$ written it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{DC}=\left(7-2\right)i+\left(7-3\right)j+\left(5-4\right)k\\ =5i+4j+k\end{array}$

It is observed that,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AD}=\stackrel{\rightharpoonup }{BC}\\ \stackrel{\rightharpoonup }{AB}=\stackrel{\rightharpoonup }{DC}\end{array}$

It can be said that, $\stackrel{\rightharpoonup }{AD}$ is parallel to $\stackrel{\rightharpoonup }{BC}$ and $\stackrel{\rightharpoonup }{AB}$ is parallel to $\stackrel{\rightharpoonup }{DC}$ .

Now, side $\stackrel{\rightharpoonup }{AB}$ is adjacent to side $\stackrel{\rightharpoonup }{AD}$ . So, area of parallelogram is, $\Vert \stackrel{\rightharpoonup }{AB}\times \stackrel{\rightharpoonup }{AD}\Vert$ .

Recall that if $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ are the adjacent sides of a parallelogram then area of the parallelogram is $\Vert a\times b\Vert$ .

Also, if there are two three-dimensional vectors in a space given in the form of their components, say, $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ , then the cross product of two vectors is defined as ,

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

Compute the value, $\stackrel{\rightharpoonup }{AB}\times \stackrel{\rightharpoonup }{AD}$ ,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AB}\times \stackrel{\rightharpoonup }{AD}=\begin{array}{ccc}i& j& k\\ 5& 4& 1\\ 1& 2& 3\end{array}\\ =\left(12-2\right)i-\left(15-1\right)j+\left(10-4\right)k\\ \text{=}10i-14j+6k\end{array}$

And, magnitude of $\stackrel{\rightharpoonup }{AB}\times \stackrel{\rightharpoonup }{AD}$ is,

  $\begin{array}{c}\Vert \stackrel{\rightharpoonup }{AB}\times \stackrel{\rightharpoonup }{AD}\Vert =\sqrt{{10}^2+{\left(-14\right)}^2+6^2}\\ \text{=}2\sqrt{83}\end{array}$

Thus, the area of the parallelogramis $2\sqrt{83}$ .