Chapter 10.4, Problem 61E

To determine

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

Answer to Problem 61E

The angle between the two adjacent sides of the tank is $90°$ .

Explanation of Solution

Given information:

A company that produces after-market machine parts is designing a reproduction of a fuel tank with the shape and dimensions shown in the figure.

  

Calculation:

The one side of the tank is the plane passing through the points $\left(0,0,0\right)$ , $\left(2,2,12\right)$ and $\left(8,8,12\right)$ .

Find the component form of the vector $u$ with initial point $\left(0,0,0\right)$ and terminal point $\left(2,2,12\right)$ .

  $\begin{array}{c}u=\langle 2-0,2-0,12-0\rangle \\ =\langle 2,2,12\rangle \end{array}$

Now find the component form of the vector $v$ with initial point $\left(0,0,0\right)$ and terminal point $\left(8,8,12\right)$ .

  $\begin{array}{c}v=\langle 8-0,8-0,12-0\rangle \\ =\langle 8,8,12\rangle \end{array}$

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

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 2& 2& 12\\ 8& 8& 12\end{array}\right|\\ =\text{i}\left(12\times 2-8\times 12\right)-\text{j}\left(12\times 2-8\times 12\right)+\text{k}\left(8\times 2-8\times 2\right)\\ =\text{i}\left(24-96\right)-\text{j}\left(24-96\right)+\text{k}\left(16-16\right)\\ =-72\text{i}+\text{72j}\end{array}$

Use the general equation of the plane.

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

Substitute $-72$ , $72$ , $0$ for $a$ , $b$ , $c$ and $0$ , $0$ , $0$ for $x_1$ , $y_1$ , $z_1$ respectively in the equation of plane.

  $\begin{array}{c}a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0\\ -72\left(x-0\right)+72\left(y-0\right)+0\left(z-0\right)=0\\ -72x+72y=0\\ x-y=0\end{array}$

So, the general equation of the plane of the first side of the tank is $x-y=0$ .

The one side of the tank is the plane passing through the points $\left(0,0,0\right)$ , $\left(10,0,0\right)$ and $\left(10,10,0\right)$ .

First find the component form of the vector $u$ with initial point $\left(0,0,0\right)$ and terminal point $\left(10,0,0\right)$ .

  $\begin{array}{c}u=\langle 10-0,0-0,0-0\rangle \\ =\langle 10,0,0\rangle \end{array}$

Now find the component form of the vector $v$ with initial point $\left(0,0,0\right)$ and terminal point $\left(10,10,0\right)$ .

  $\begin{array}{c}v=\langle 10-0,10-0,0-0\rangle \\ =\langle 10,10,0\rangle \end{array}$

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

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}\text{i}& \text{j}& \text{k}\\ 10& 0& 0\\ 10& 10& 0\end{array}\right|\\ =\text{i}\left(0\times 0-10\times 0\right)-\text{j}\left(0\times 10-10\times 0\right)+\text{k}\left(10\times 10-10\times 0\right)\\ =\text{i}\left(0-0\right)-\text{j}\left(0-0\right)+\text{k}\left(100-0\right)\\ =100\text{k}\end{array}$

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 $0$ , $0$ , $100$ for $a$ , $b$ , $c$ and $0$ , $0$ , $0$ for $x_1$ , $y_1$ , $z_1$ respectively in the equation of plane.

  $\begin{array}{c}a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0\\ 0\left(x-0\right)+0\left(y-0\right)+100\left(z-0\right)=0\\ 100z=0\\ z=0\end{array}$

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 $x-y=0$ is $\langle 1,-1,0\rangle$ 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 $\langle 1,-1,0\rangle$ and $\langle 0,0,1\rangle$ .

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

Now substitute all the values in the formula for angle.

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

The solution for the trigonometric equation $\cos \theta =0$ is $\theta =90°$ .

Therefore, the angle between the two adjacent sides of the tank is $90°$ .