Chapter 12, Problem 28CLT

To determine

Find the angle.

Answer to Problem 28CLT

  $\boxed{\theta =84.26°}$

Explanation of Solution

Given information:

A plastic wastebasket has the shape and dimensions shown in the figure. In fabricating a mold for making the wastebasket, it is necessary to know the angle between two adjacent sides. Find the angle.

  

Calculation:

Consider the given figure,

  

Now from the figure tabulate the coordinates of vertices.

  $\begin{array}{ccccc}side& vertex1& vertex2& vertex3& vertex4\\ side1& \left(2,0,0\right)& \left(2,2,0\right)& \left(3,3,3\right)& \left(3,-1,3\right)\\ side2& \left(2,0,0\right)& \left(0,0,0,\right)& \left(-1,-1,3\right)& \left(3,-1,3\right)\\ side3& \left(0,2,0,\right)& \left(0,0,0,\right)& \left(-1,-1,3\right)& \left(-1,3,3\right)\\ side4& \left(0,2,0,\right)& \left(2,2,0\right)& \left(3,3,3\right)& \left(-1,3,3\right)\\ side5& \left(0,2,0,\right)& \left(2,2,0\right)& \left(0,0,0\right)& \left(2,0,0\right)\end{array}$

Now calculate normal vectors of sides.

Normal vector of side $1$ follow the procedure, find the diagonal vector from their position vector of the end point using standard results.

For the two position vector,

The vector joining term is,

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

Diagonal vector joining $\left(3,3,3\right)\&\left(2,0,0\right)$

  $\begin{array}{l}U=\left(3,3,3\right)-\left(2,0,0\right)\\ U=\left(1,3,3\right)\end{array}$

Diagonal vector joining $\left(3,-1,3\right)\&\left(2,2,0\right)$

  $\begin{array}{l}V=\left(3,-1,3\right)-\left(2,2,0\right)\\ V=\left(1,-3,3\right)\end{array}$

Now find cross product of diagonal vectors.

  $\begin{array}{l}U=\langle a,b,c\rangle \\ V=\langle d,e,f\rangle \end{array}$

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

Now,

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

Hence normal vector of side $1$ is $\boxed{n_1=18i-6k}$

Normal vector of side $2$

Diagonal vector joining $\left(2,0,0\right)\&\left(-1,-1,3\right)$

  $\begin{array}{l}U=\left(2,0,0\right)-\left(-1,-1,3\right)\\ U=\left(3,1,-3\right)\end{array}$

Diagonal vector joining $\left(3,-1,3\right)\&\left(0,0,0\right)$

  $\begin{array}{l}V=\left(3,-1,3\right)-\left(0,0,0\right)\\ V=\left(3,-1,3\right)\end{array}$

Now cross product of diagonal vectors.

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

Hence normal vector of side $2$ is $\boxed{-18j-6k}$

Now calculate angle between two vectors by standard formula.

Angle between two vectors $n_1\&n_2$ is,

  $\begin{array}{l}\cos \theta =\frac{n_1\cdot n_2}{\left|n_1\right|\left|n_2\right|}\\ n_1\cdot n_2=\langle 18,0,-6\rangle \cdot \langle 0,-18,-6\rangle \\ =\langle 0+0+36\rangle \\ =36\\ \left|n_1\right|=\sqrt{{18}^2+6^2}\\ =\sqrt{360}\\ \left|n_2\right|=\sqrt{{18}^2+6^2}\\ =\sqrt{360}\\ \cos \theta =\frac{36}{\left(\sqrt{360}\right)\left(\sqrt{360}\right)}\\ =\frac{1}{10}\\ \theta ={\cos }^{-1}\left(0.1\right)\\ \theta =84.26°\end{array}$

Hence angle between two adjacent sides is $\boxed{\theta =84.26°}$