Chapter 10.4, Problem 62E

To determine

To find: The angle between the two adjacent sides of the pan.

Answer to Problem 62E

The angle between the two adjacent sides of the pan is $86.34°$ .

Explanation of Solution

Given information:

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

  

Calculation:

The one side of the pan is the plane passing through the points $\left(0,0,0\right)$ , $\left(-1,-1,7\right)$ and $\left(9,19,7\right)$ .

First find the component form of the vector $u$ with initial point $\left(0,0,0\right)$ and terminal point $\left(-1,-1,7\right)$ .

  $\begin{array}{c}u=\langle -1-0,-1-0,7-0\rangle \\ =\langle -1,-1,7\rangle \end{array}$

Now find the component form of the vector $v$ with initial point $\left(0,0,0\right)$ and terminal point $\left(9,19,7\right)$ .

  $\begin{array}{c}v=\langle 9-0,19-0,7-0\rangle \\ =\langle 9,19,7\rangle \end{array}$

Use the cross product of vectors $u$ and $v$ for the normal vector to the plane.

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ -1& -1& 7\\ 9& 19& 7\end{array}\right|\\ =\text{i}\left(7\times \left(-1\right)-19\times 7\right)-\text{j}\left(7\times -1-9\times 7\right)+\text{k}\left(-1\times 19-\left(-1\right)\times 9\right)\\ =\text{i}\left(-7-133\right)-\text{j}\left(-7-63\right)+\text{k}\left(-19+9\right)\\ =-140\text{i}+70\text{j}-10\text{k}\end{array}$

Use the general equation of the plane.

  $a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$

Now substitute $-140$ , $70$ , $-10$ for $a$ , $b$ , $c$ and $0$ , $0$ , $0$ for $x_1$ , $y_1$ , $z_1$ respectively in the equation of plane.

  $\begin{array}{c}a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0\\ -140\left(x-0\right)+70\left(y-0\right)+\left(-10\right)\left(z-0\right)=0\\ -140x+70y-10z=0\\ 14x-7y+z=0\end{array}$

So, The general equation of the plane of the first side of the tank is $14x-7y+z=0$ .

The second side of the pan is the plane passing through the points $\left(0,0,0\right)$ , $\left(8,0,0\right)$ and $\left(8,18,0\right)$ .

First find the component form of the vector $u$ with initial point $\left(0,0,0\right)$ and terminal point $\left(8,0,0\right)$ .

  $\begin{array}{c}u=\langle 8-0,0-0,0-0\rangle \\ =\langle 8,0,0\rangle \end{array}$

Now find the component form of the vector $v$ with initial point $\left(0,0,0\right)$ and terminal point $\left(8,18,0\right)$ .

  $\begin{array}{c}v=\langle 8-0,18-0,0-0\rangle \\ =\langle 8,18,0\rangle \end{array}$

Use the cross product of vectors $u$ and $v$ for the normal vector to the plane.

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 8& 0& 0\\ 8& 18& 0\end{array}\right|\\ =\text{i}\left(0\times 0-18\times 0\right)-\text{j}\left(0\times 8-8\times 0\right)+\text{k}\left(18\times 8-8\times 0\right)\\ =\text{i}\left(0-0\right)-\text{j}\left(0-0\right)+\text{k}\left(144-0\right)\\ =144\text{k}\end{array}$

Use the general equation of the plane.

  $a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0$

Now substitute $0$ , $0$ , $144$ for $a$ , $b$ , $c$ and $0$ , $0$ , $0$ for $x_1$ , $y_1$ , $z_1$ respectively in the equation of plane.

  $\begin{array}{c}a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0\\ 0\left(x-0\right)+0\left(y-0\right)+144\left(z-0\right)=0\\ 144z=0\\ z=0\end{array}$

So, The general equation of the plane of the first side of the tank is $z=0$ .

Compare the equation of the plane with $ax+by+cz=d$ .

So, the normal vector for plane $14x-7y+z=0$ is $\langle 14,-7,1\rangle$ and normal vector for plane $z=0$ is $\langle 0,0,1\rangle$ .

The angle between the two planes is equal to the angles between their normal vectors.

Use the formula for angle between two vectors $u$ and $v$ .

  $\cos \theta =\frac{u\cdot v}{\Vert u\Vert \Vert v\Vert }$

The dot product of both the vectors $\langle 14,-7,1\rangle$ and $\langle 0,0,1\rangle$ .

  $\begin{array}{c}\langle 14,-7,1\rangle \cdot \langle 0,0,1\rangle =14\times 0+\left(-7\right)\times 0+1\times 1\\ =1\end{array}$

The magnitude of the vector $\langle 14,-7,1\rangle$ .

  $\begin{array}{c}\Vert \langle 14,-7,1\rangle \Vert =\sqrt{{\left(14\right)}^2+{\left(-7\right)}^2+{\left(1\right)}^2}\\ =\sqrt{196+49+1}\\ =\sqrt{246}\end{array}$

The magnitude of the vector $\langle 0,0,1\rangle$ .

  $\begin{array}{c}\Vert \langle 0,0,1\rangle \Vert =\sqrt{{\left(0\right)}^2+{\left(0\right)}^2+{\left(1\right)}^2}\\ =\sqrt{0+0+1}\\ =1\end{array}$

Now substitute all the values in the formula for angle.

  $\begin{array}{c}\cos \theta =\frac{\langle 14,-7,1\rangle \cdot \langle 0,0,1\rangle }{\Vert \langle 14,-7,1\rangle \Vert \Vert \langle 0,0,1\rangle \Vert }\\ =\frac{1}{\sqrt{246}\times 1}\\ =\frac{\sqrt{246}}{246}\end{array}$

The solution for the trigonometric equation $\cos \theta =\frac{\sqrt{246}}{246}$ is $\theta =86.34°$ .

Therefore, he atngle between the two adjacent sides of the pan is $86.34°$ .