Chapter 11.4, Problem 68E

To determine

Find the angle between two adjacent sides of the pan.

Answer to Problem 68E

  $\boxed{\theta =88.9°}$

Explanation of Solution

Given information:

A bread pan is tapered so that a loaf of bread can be removed easily. The shape and dimensions of the pan are shown in the figure. Find the angle between two adjacent sides of the pan.

  

Calculation:

Consider the vector in space $a=\left(x_1,y_1,z_1\right)$ and $b=\left(x_2,y_2,z_2\right)$ so the cross product of , $a=\left(x_1,y_1,z_1\right)$ and $b=\left(x_2,y_2,z_2\right)$ will be,

  $\begin{array}{l}a\times b=\left|\begin{array}{ccc}i& j& k\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ =i\left[y_1\times z_2-y_2\times z_1\right]-j\left[x{}_1\times z_2-x_2{z}_1\right]+k\left[x_1\times y_2-x_2\times y_1\right]\end{array}$

Angle between $a=\left(x_1,y_1,z_1\right)$ and $b=\left(x_2,y_2,z_2\right)$ will be,

  $\Vert a\times b\Vert =\Vert a\Vert \Vert b\Vert \sin \theta$

Now according to the figure,

  

To obtain a normal vector $n_1$ for the first side of the pan having points, $P,S,T$ . Now find the cross product of $\overrightarrow{SP}=\left(8,0,0\right)$ and $\overrightarrow{ST}=\left(-1,-1,7\right)$

  $\begin{array}{l}{n}_1=\left|\begin{array}{ccc}i& j& k\\ 8& 0& 0\\ -1& -1& 7\end{array}\right|\\ =i\left(0-0\right)-j\left(56-0\right)+k\left(-8-0\right)\\ =-56j-8k\end{array}$

Now a normal vector $n_2$ for the second side of the pan having points, $U,S,T$ Now find the cross product of $\overrightarrow{SU}=\left(0,18,0\right)$ and $\overrightarrow{ST}=\left(-1,-1,7\right)$

  $\begin{array}{l}{n}_2=\left|\begin{array}{ccc}i& j& k\\ 0& 18& 0\\ -1& -1& 7\end{array}\right|\\ =i\left(126-0\right)-j\left(0-0\right)+k\left(0+18\right)\\ =126i+18k\end{array}$

Now the normal vector of the plane are $n_1=\left(0,-56,-8\right)$ and $n_2=\left(126,0,18\right)$ so angle between two sides be find as,

  $\begin{array}{l}\cos \theta =\left|\frac{n_1.n_2}{\left|n_1\right|\times \left|n_2\right|}\right|\\ =\left|\frac{\left(0,-56,-8\right)\cdot \left(126,0,18\right)}{\left|\left(0,-56,-8\right)\left|\left(126,0,18\right)\right|\right|}\right|\\ =\left|\frac{\left(0,-56,-8\right)\cdot \left(126,0,18\right)}{\sqrt{0^2+{\left(-56\right)}^2+{\left(-8\right)}^2}\times \sqrt{{126}^2+0^2+{18}^2}}\right|\\ \cos \theta =\frac{144}{\sqrt{3200}\sqrt{16200}}\end{array}$

  $\begin{array}{l}\theta ={\cos }^{-1}\left(\frac{144}{\sqrt{3200}\sqrt{16200}}\right)\\ ={\cos }^{-1}\left(\frac{144}{7200}\right)\\ \boxed{\theta =88.9°}\end{array}$

Hence, between two adjacent sides $\boxed{\theta =88.9°}$ .