Chapter 11.3, Problem 43E

$\left(a\right)$

To determine

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

Explanation of Solution

Given information:

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

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 $A\left(3,2-1\right),B\left(-2,2,-3\right),C\left(3,5,-2\right),D\left(-2,5,-4\right)$ .

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 $\stackrel{\rightharpoonup }{AB}$ is evaluated as, subtract the coordinate of B from A and write it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AB}=\left(-2-3\right)i+\left(2-2\right)j+\left(-3+1\right)k\\ =-5i+0j-2k\end{array}$

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

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

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

  $\begin{array}{c}\stackrel{\rightharpoonup }{CD}=\left(-2-3\right)i+\left(5-5\right)j+\left(-4+2\right)k\\ =-5i+0j-2k\end{array}$

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

  $\begin{array}{c}\stackrel{\rightharpoonup }{DA}=\left(3+2\right)i+\left(2-5\right)j+\left(-1+4\right)k\\ =5i-3j+3k\end{array}$

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

  $\begin{array}{c}\stackrel{\rightharpoonup }{AC}=\left(3-3\right)i+\left(5-2\right)j+\left(-2+1\right)k\\ =0i+3j-k\end{array}$

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

  $\begin{array}{c}\stackrel{\rightharpoonup }{BD}=\left(-2+2\right)i+\left(5-2\right)j+\left(-4+3\right)k\\ =0i+3j-k\end{array}$

It is observed that,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AB}=\stackrel{\rightharpoonup }{CD}\\ \stackrel{\rightharpoonup }{AC}=\stackrel{\rightharpoonup }{BD}\end{array}$

It can be said that, $\stackrel{\rightharpoonup }{AB}$ is parallel to $\stackrel{\rightharpoonup }{CD}$ and $\stackrel{\rightharpoonup }{AC}$ is parallel to $\stackrel{\rightharpoonup }{BD}$ .

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

Hence, verified.

$\left(b\right)$

To determine

To calculate:The area of the parallelogram.

Answer to Problem 43E

The area of the parallelogramis $\sqrt{286}$ .

Explanation of Solution

Given information:

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

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 $A\left(3,2-1\right),B\left(-2,2,-3\right),C\left(3,5,-2\right),D\left(-2,5,-4\right)$ .

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 $\stackrel{\rightharpoonup }{AB}$ is evaluated as, subtract the coordinate of B from A and write it in vector notation as,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AB}=\left(-2-3\right)i+\left(2-2\right)j+\left(-3+1\right)k\\ =-5i+0j-2k\end{array}$

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

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

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

  $\begin{array}{c}\stackrel{\rightharpoonup }{CD}=\left(-2-3\right)i+\left(5-5\right)j+\left(-4+2\right)k\\ =-5i+0j-2k\end{array}$

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

  $\begin{array}{c}\stackrel{\rightharpoonup }{DA}=\left(3+2\right)i+\left(2-5\right)j+\left(-1+4\right)k\\ =5i-3j+3k\end{array}$

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

  $\begin{array}{c}\stackrel{\rightharpoonup }{AC}=\left(3-3\right)i+\left(5-2\right)j+\left(-2+1\right)k\\ =0i+3j-k\end{array}$

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

  $\begin{array}{c}\stackrel{\rightharpoonup }{BD}=\left(-2+2\right)i+\left(5-2\right)j+\left(-4+3\right)k\\ =0i+3j-k\end{array}$

It is observed that,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AB}=\stackrel{\rightharpoonup }{CD}\\ \stackrel{\rightharpoonup }{AC}=\stackrel{\rightharpoonup }{BD}\end{array}$

It can be said that, $\stackrel{\rightharpoonup }{AB}$ is parallel to $\stackrel{\rightharpoonup }{CD}$ and $\stackrel{\rightharpoonup }{AC}$ is parallel to $\stackrel{\rightharpoonup }{BD}$ .

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

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, $\stackrel{\rightharpoonup }{AC}\times \stackrel{\rightharpoonup }{AB}$ ,

  $\begin{array}{c}\stackrel{\rightharpoonup }{AC}\times \stackrel{\rightharpoonup }{AB}=\begin{array}{ccc}i& j& k\\ 0& 3& -1\\ -5& 0& -2\end{array}\\ =\left(-6\right)i-\left(-5\right)j+\left(15\right)k\\ \text{=}-6i+5j+15k\end{array}$

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

  $\begin{array}{c}\Vert \stackrel{\rightharpoonup }{AC}\times \stackrel{\rightharpoonup }{AB}\Vert =\sqrt{{\left(-6\right)}^2+{\left(5\right)}^2+{\left(15\right)}^2}\\ \text{=}\sqrt{286}\end{array}$

Thus, the area of the parallelogramis $\sqrt{286}$ .