How to find the angle between 2 intersection lines of four planes?

Question

Given $L_1=\begin{cases}P_1: x-y+z=4\\ P_2: 2x+y-z=6 \end{cases}$

and $L_2=\begin{cases}P_1:x+y+z=4\\ P_2:2x+3y-z=6 \end{cases}$

$L_1$ and $L_2$ are the intersection lines of the given planes. Find the angle between the lines $L_1,L_2$

First in order to find the intersection lines between the respective planes it is enough to find to dots that belong to the intersection line, so for $L_1$: $$ P_1+P_2 \Rightarrow 3x=10 \Rightarrow x={10 \over 3} $$

Then: $$ 2 \cdot \frac{10}{3}+y-z=6 \Rightarrow y-z=-\frac{2}{3} $$ For $z=0$ and $z=1$ we get $y=-\frac{2}{3}$ and $y=\frac{1}{3}$ respectively so now can represent $L_1$ parametrically: $(\frac{10}{3},\frac{1}{3}, 10)+t(0,-1,0)$ where $t(0,-1,0)$ is the direction vector.

After identical calculations we can find $L_2: (0,\frac{10}{4}, \frac{3}{2})+t(\frac{10}{3}, 0,-\frac{2}{3})$.

Now we can calculate the angle of $L_1, L_2$ via scalar product: $$ \cos \theta=\frac{L_1 \cdot L_2}{\Vert{L_1}\Vert \cdot \Vert L_2 \Vert}=0 \Rightarrow \theta=90^{\circ} $$

I'm not sure that my answer is correct and I feel I'm missing something.

Answer 1

The idea to find direction vectors of both lines is fine, but it's a bit unclear to me how you (think you) arrive at these direction vectors...

You can either 'solve' both systems and from the parametric form of the two solution sets, you can simply read off the two direction vectors.

Alternatively, note that you can also simply read off the normal vectors of both planes in each system and the cross product of these normal vectors give you the respective direction vectors, so: $$\left( 1,-1,1 \right) \times \left( 2,1,-1 \right) = \left( 0,3,3 \right) \quad\mbox{and}\quad \left( 1,1,1 \right) \times \left( 2,3,-1 \right) = \left( -4,3,1 \right)$$ So you're looking for the angle between $\left( 0,3,3 \right)$ (or $\left( 0,1,1 \right)$, since you can scale!) and $\left( -4,3,1 \right)$.

Answer 2

Let $\vec{l_1}(a,b,c)$.

Thus, $a-b+c=0$ and $2a+b-c=0$, which gives $a=0$ and $b=c$ and $\vec{l_1}(0,1,1).$

Let $\vec{l_2}(a,b,c)$.

Thus, $a+b+c=0$ and $2a+3b-c=0$, which gives $b=-\frac{3}{4}a$, $c=-\frac{1}{4}a$

and $\vec{l_1}(4,-3,-1).$

Now, $$\measuredangle\left(L_1,L_2\right)=\arccos\frac{|0\cdot4+1\cdot(-3)+1\cdot(-1)|}{\sqrt{4^2+(-3)^2+(-1)^2}\sqrt{0^2+1^2+1^2}}=\arccos\frac{2}{\sqrt{13}}.$$

Answer 3

Let me answer the general case. First finding intersection of planes: In general planes in $\mathbb{R}^3$ are of the form: $$ax+by+cz+d=0.$$ The intersection of two planes $ax+by+cz+d=0$ and $Ax+By+Cz+D=0$ exists if and only if $|aA+bB+cC|\neq ||(a,b,c)||.||(A,B,C)||$ or if $|aA+bB+cC|=||(a,b,c)||.||(A,B,C)||$, then $dC=Dc$. The latter case means planes coincide. Note that $|aA+bB+cC|=||(a,b,c)||.||(A,B,C)||$ implies two planes are parallel. If two planes have intersection then by letting one of the variables $x$, $y$ or $z$ equal to $t$ we will find other variables. Here let $z=t$ then solve: $$\left\{\begin{array}{} ax+by=d-ct\\ Ax+By=D-Ct\end{array}\right.$$ we are now able to find values of $x$ and $y$. After calculation we have something like the following form: $$\frac{x-a'}{a''}=\frac{y-b'}{b''}=\frac{z}{1}(=t).$$ Now, do the similar arguments for two other planes. We have: $$\frac{x-A'}{A''}=\frac{y-B'}{B''}=\frac{z}{1}(=T).$$ Now we have two vectors $(A'',B'',1)$ and $(a'',b'',1)$ and by using the formula: $a''A''+b''B''+1=\sqrt{a''^2+b''^2+1}.\sqrt{A''^2+B''^2+1}.cos(\alpha)$ the angle $\alpha$ is easy to obtain.

Now in your question: $$L_1: x=\frac{10}{3},y+\frac{2}{3}=z(=t)$$ here $v_1=(0,1,1)$ and $$L_2:\frac{x-6}{-4}=\frac{y+2}{3}=z(=t)$$ here $w=(-4,3,1)$. So, $$cos(\alpha)=\frac{4}{\sqrt{2}\sqrt{26}}=\frac{2\sqrt{13}}{13}$$