Chapter 11.4, Problem 67E

To determine

Find the angle between two adjacent sides.

Answer to Problem 67E

  $\boxed{\theta =82.93°}$

Explanation of Solution

Given information:

A chute at the top of a grain elevator of a combine funnels the grain into a bin, as shown in the figure. Find the angle between two adjacent sides.

  

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,

  

The vector of point $P\left(6,0,0\right)and,S\left(0,0,0\right)$ is $\left(6,0,0\right)$ and the vector of points, $T\left(-1,-1,8\right)and,S\left(0,0,0\right)$ is $\left(-1,-1,8\right)$ .

Consider $ST=a$ = $\left(-1,-1,8\right)$

  $SP=b=$

  $\left(6,0,0\right)$

Now find angle between two vectors $a$ and $b$ will be,

  $\begin{array}{l}\cos \theta =\left|\frac{a.b}{\left|a\right|\times \left|b\right|}\right|\\ =\left|\frac{\left(6,0,0\right)\cdot \left(-1,-1,8\right)}{\sqrt{6^2+0+0}\times \sqrt{1^2+1^2+8^2}}\right|\\ =\left|\frac{-6+0+0}{6\times \sqrt{66}}\right|\\ \cos \theta =\frac{1}{\sqrt{66}}\\ \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{66}}\right)\\ \theta =82.93°\end{array}$

Hence between two adjacent sides $\boxed{\theta =82.93°}$ .