Chapter 11, Problem 18CLT

To determine

To Calculate:

The angle of plastic wastebasket.

Answer to Problem 18CLT

The angle is ${84.26}^{\circ }$ .

Explanation of Solution

Given Information:

  

Calculation:

To get the angle between the adjacent sides, proceed as follows:

Get the coordinates of the vertices of endpoints of the adjacent sides.

Side	Vertex $1$	Vertex $2$	Vertex $3$	Vertex $4$
Side $1$	$\left(2,0,0\right)$	$\left(2,2,0\right)$	$\left(3,3,3\right)$	$\left(3,-1,3\right)$
Side $2$	$\left(2,0,0\right)$	$\left(0,0,0\right)$	$\left(-1,-1,3\right)$	$\left(3,-1,3\right)$
Side $3$	$\left(0,2,0\right)$	$\left(0,0,0\right)$	$\left(-1,-1,3\right)$	$\left(-1,3,3\right)$
Side $4$	$\left(0,2,0\right)$	$\left(2,2,0\right)$	$\left(3,3,3\right)$	$\left(-1,3,3\right)$
Side $5$	$\left(0,2,0\right)$	$\left(2,2,0\right)$	$\left(0,0,0\right)$	$\left(2,0,0\right)$

Now to get the normal vectors of the sides of the figure:

Normal vector of side $1$ :

To determine the normal vector of the side procced as follows:

Get the diagonal vectors from their position vectors of the endpoints using the standard result.

Standard result:

For the two position vectors:

  $\begin{array}{l}\text{W=}\langle a,b,c\rangle \\ \text{T=}\langle d,e,f\rangle \end{array}$

The vector joining them is,

  $\text{W-T=}\langle a,b,c\rangle -\langle d,e,f\rangle$

Diagonal vectors joining $\langle 3,3,3\rangle$ and $\langle 2,0,0\rangle$ is

  $\begin{array}{l}\text{U}=\langle 3,3,3\rangle -\langle 2,0,0\rangle \\ \text{U}=\langle 1,3,3\rangle \end{array}$

Diagonal vectors joining $\langle 3,-1,3\rangle$ and $\langle 2,2,0\rangle$ is

  $\begin{array}{l}\text{V}=\langle 3,-1,3\rangle -\langle 2,2,0\rangle \\ \text{V}=\langle 1,-3,3\rangle \end{array}$

Now, to get cross product between the diagonal vectors using standard result.

Standard result:

For the vectors:

  $\begin{array}{l}\text{U}=\langle a,b,c\rangle \\ \text{V}=\langle d,e,f\rangle \\ \text{U}\times \text{V}=\left|\begin{array}{ccc}i& j& k\\ a& b& c\\ d& e& f\end{array}\right|\end{array}$

Therefore,

  $\begin{array}{l}\text{U}\times \text{V}=\left|\begin{array}{ccc}i& j& k\\ 1& 3& 3\\ 1& -3& 3\end{array}\right|\\ =\left(9+9\right)i-\left(3-3\right)j+\left(-3-3\right)k\\ =18i-6k\end{array}$

Hence, the normal vector of the side $1$ is $n_1=18i-6k$ .

Now to get the normal vector to the side $2$

Diagonal vectors joining $\langle 2,0,0\rangle$ and $\langle -1,-1,3\rangle$ is

  $\begin{array}{l}\text{U}=\langle 2,0,0\rangle -\langle -1,-1,3\rangle \\ \text{U}=\langle 3,1,-3\rangle \end{array}$

Diagonal vectors joining $\langle 3,-1,3\rangle$ and $\langle 0,0,0\rangle$ is

  $\begin{array}{l}\text{V}=\langle 3,-1,3\rangle -\langle 0,0,0\rangle \\ \text{V}=\langle 3,-1,3\rangle \end{array}$

Now get the cross product of two diagonal vectors.

  $\begin{array}{l}\text{U}\times \text{V}=\left|\begin{array}{ccc}i& j& k\\ 3& 1& -3\\ 3& -1& 3\end{array}\right|\\ =\left(3-3\right)i-\left(9+9\right)j+\left(-3-3\right)k\\ =-18i-6k\end{array}$

Hence, the normal vector to the side $2$ is $n_2=-18i-6k$

Now, get the angle between the two vectors using the standard result.

Angle between two vectors $n_1,n_2$ is $\cos \theta =\frac{n_1\cdot n_2}{\left|n_1\right|\left|n_2\right|}$

  $\begin{array}{l}{n}_1\cdot n_2=\langle 18,0,-6\rangle \cdot \langle 0,-18,-6\rangle \\ =\left(0+0+36\right)\\ =36\\ \left|n_1\right|=\sqrt{{18}^2+6^2}\\ =\sqrt{360}\\ \left|n_2\right|=\sqrt{{18}^2+6^2}\\ =\sqrt{360}\end{array}$

Hence the angle is

  $\begin{array}{l}\cos \theta =\frac{36}{\left(\sqrt{360}\right)\left(\sqrt{360}\right)}\\ \cos \theta =\frac{1}{10}\\ \theta ={\cos }^{-1}\left(0.1\right)\\ \theta ={84.26}^{\circ }\end{array}$

Hence the angle between the two sides is ${84.26}^{\circ }$ .