Chapter 11.4, Problem 60E

To determine

Find the angle $\theta$ between the two adjacent sides of the pain.

Answer to Problem 60E

The angle between two adjacent side is $\theta \approx {91.15}^{°}$

Explanation of Solution

Given information:

We are given the points: $T\left(-1,-1,7\right),R\left(9,19,7\right),P\left(8,0,0\right),Q\left(8,18,0\right),S\left(0,0,0\right)$

Formula used:

  $\cos \theta =\frac{\left|n_1\cdot n_2\right|}{\Vert n_1\Vert \Vert n_2\Vert }$

The angle between the planes is $\cos \theta =\frac{\left|n_1\cdot n_2\right|}{\Vert n_1\Vert \Vert n_2\Vert }$ .

Now calculating the vector $\overrightarrow{PQ}$ :

  $\overrightarrow{PQ}=\left(8-8\right)i+\left(18-0\right)j+\left(0-0\right)k=18j$

Now calculating the vector $\overrightarrow{RQ}$ :

  $\overrightarrow{RQ}=\left(8-9\right)i+\left(18-19\right)j+\left(0-7\right)k=-i-j-7k$

The normal $n_1$ on the plane containing $\overrightarrow{PQ}$ and $\overrightarrow{RQ}$ is:

  $\begin{array}{l}{n}_1=\overrightarrow{PQ}\times \overrightarrow{RQ}=\left|\begin{array}{ccc}i& j& k\\ 0& 18& 0\\ -1& -1& -7\end{array}\right|\\ \\ =i\left|\begin{array}{cc}18& 0\\ -1& -7\end{array}\right|-j\left|\begin{array}{cc}0& 0\\ -1& -7\end{array}\right|+k\left|\begin{array}{cc}0& 18\\ -1& -1\end{array}\right|\\ \\ =i\left[18\left(-7\right)-0\left(-1\right)\right]-j\left[0\left(-7\right)-0\left(-1\right)\right]+k\left[0\left(-1\right)-18\left(-1\right)\right]\\ \\ =-126i+18k\end{array}$

The vector $\overrightarrow{AQ}$ is:

  $\begin{array}{l}A\left(0,18,0\right)\\ \overrightarrow{AQ}=\left(8-0\right)i+\left(18-18\right)j+\left(0-0\right)k=8i\end{array}$

The normal $n_2$ on the plane containing $\overrightarrow{AQ}$ and $\overrightarrow{RQ}$ is:

  $\begin{array}{l}{n}_2=\overrightarrow{AQ}\times \overrightarrow{RQ}=\left|\begin{array}{ccc}i& j& k\\ 8& 0& 0\\ -1& -1& -7\end{array}\right|\\ \\ =i\left|\begin{array}{cc}0& 0\\ -1& -7\end{array}\right|-j\left|\begin{array}{cc}8& 0\\ -1& -7\end{array}\right|+k\left|\begin{array}{cc}8& 0\\ -1& -1\end{array}\right|\\ \\ =i\left[0\left(-7\right)-0\left(-1\right)\right]-j\left[8\left(-7\right)-0\left(-1\right)\right]+k\left[8\left(-1\right)-0\left(-1\right)\right]\\ \\ =56j-8k\end{array}$

We determine $\cos \theta$ , where $\theta$ is the angle between the two planes:

  $\begin{array}{l}\cos \theta =\frac{\left|n_1\cdot n_2\right|}{\Vert n_1\Vert \Vert n_2\Vert }\\ \\ \cos \theta =\frac{\left|<-126,0,18>\cdot <0,56,-8>\right|}{\sqrt{{\left(126\right)}^2+{18}^2}\cdot \sqrt{{56}^2+{\left(-8\right)}^2}}\\ \\ \cos \theta =\frac{\left|-126\left(0\right)+0\left(56\right)+18\left(-8\right)\right|}{7200}\approx -0.02\\ \\ \boxed{\theta \approx {91.15}^{\circ }}\end{array}$