Chapter 11.4, Problem 59E

To determine

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

Answer to Problem 59E

The angle between two adjacent side is $\theta ={89.1}^{°}$

Explanation of Solution

Given information:

Coordinates of the given figure are $T\left(-1,-1,8\right),R\left(7,7,8\right),P\left(6,0,0\right),Q\left(6,6,0\right),S\left(0,0,0\right)$

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

1. From the three points of the plane we can compute the normal vector. First find the component vectors as $\overrightarrow{QP}=<0,-6,0>$ and $\overrightarrow{QR}=<1,1,8>$ .
2. The normal vector to the plane is given by the cross product of two component vector as $n_1=\overrightarrow{QP}\times \overrightarrow{QP}=-48i+6k=<-48,0,6>$
3. From the three points of the adjacent planes we can compute the normal vector. First find the component vector as $\overrightarrow{QR}=<1,1,8>$ and $\overrightarrow{QA}=<-6,0,0>$ .
4. The normal vector to the plane is given by the cross product of the two component vectors as $n_2=\overrightarrow{QR}\times \overrightarrow{QA}=-48j+6k=<0,-48,6>$ .
5. Substituting to the angle equation we get that:

  $\begin{array}{l}\cos \theta =\frac{36}{2340}=0.0154\\ \\ \boxed{\theta ={89.1}^{\circ }}\end{array}$