2.5 Equations of lines and planes in space  (Page 6/19)

Note that the two planes have nonparallel normals, so the planes intersect. Further, the origin satisfies each equation, so we know the line of intersection passes through the origin. Add the plane equations so we can eliminate the one of the variables, in this case, $y\text{:}$

This gives us $x=-\frac{2}{3}z.$ We substitute this value into the first equation to express $y$ in terms of $z\text{:}$

We now have the first two variables, $x$ and $y,$ in terms of the third variable, $z.$ Now we define $z$ in terms of $t.$ To eliminate the need for fractions, we choose to define the parameter $t$ as $t=-\frac{1}{3}z.$ Then, $z=\mathrm{-3}t.$ Substituting the parametric representation of $z$ back into the other two equations, we see that the parametric equations for the line of intersection are $x=2t,y=t,z=\mathrm{-3}t.$ The symmetric equations for the line are $\frac{x}{2}=y=\frac{z}{\mathrm{-3}}.$

1. The normal vectors for these planes are ${\text{n}}_1=⟨1,2,\mathrm{-1}⟩$ and ${\text{n}}_2=⟨2,4,\mathrm{-2}⟩.$ These two vectors are scalar multiples of each other. The normal vectors are parallel, so the planes are parallel.
2. The normal vectors for these planes are ${\text{n}}_1=⟨2,\mathrm{-3},2⟩$ and ${\text{n}}_2=⟨6,2,\mathrm{-3}⟩.$ Taking the dot product of these vectors, we have
   
   ${\text{n}}_1·{\text{n}}_2=⟨2,\mathrm{-3},2⟩·⟨6,2,\mathrm{-3}⟩=2(6)-3(2)+2(\mathrm{-3})=0.$
   
   The normal vectors are orthogonal, so the corresponding planes are orthogonal as well.
3. The normal vectors for these planes are ${\text{n}}_1=⟨1,1,1⟩$ and ${\text{n}}_2=⟨1,\mathrm{-3},5⟩\text{:}$
   
   $\begin{array}{cc}\hfill \text{cos}\theta & =\frac{\left|{\text{n}}_1·{\text{n}}_2\right|}{\Vert {\text{n}}_1\Vert \Vert {\text{n}}_2\Vert }\hfill \\ & =\frac{\left|⟨1,1,1⟩·⟨1,\mathrm{-3},5⟩\right|}{\sqrt{1^2+1^2+1^2}\sqrt{1^2+{(\mathrm{-3})}^2+5^2}}\hfill \\ & =\frac{3}{\sqrt{105}}.\hfill \end{array}$
   
   The angle between the two planes is $1.27$ rad, or approximately $73\text{°}.$