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

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.

• the parametric equation for $L_1$ is not correct. Have you tried using cross-product to get the directions of the lines? It might be easier... – David Quinn Apr 26 '17 at 13:07
• @DavidQuinn if I do the vector product I will just get the direction vectors right? Which is essentially all I need anyway? – Yos Apr 26 '17 at 13:09
• The second is not correct either. To solve for the equation of the first line, consider the simultenous system of two equations, introduce free parameters if any etc. Similarly you solve for the second line. Then use the formula you have. – AnyAD Apr 26 '17 at 13:10
• yes, see the answer from @StackTD below – David Quinn Apr 26 '17 at 13:11

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 $\left( -4,3,1 \right)$.

• @ StackTD: thanks for the answer, it totally makes sense but can you please point me where my problem with the calculation lies? This is my complete calculation of the parametric form for $L_1$: 1) $P_1+P_2 \Rightarrow 3x=10$ so we found the $x$. 2) Now we can plug it into $P_1$: ${10 \over 3}-y+z=4$ so $z-y={2 \over 3}$ 3) if I choose $y=0$ I get $z={2 \over 3}$, for $y=1$ I get $z={5 \over 3}$. So now I have $({10 \over 3},0,{2 \over 3})+t({10 \over 3}, 1, {5 \over 3})$ but it's still not correct... – Yos Apr 26 '17 at 13:22
• @Yos Picking $y=0$ gives you a single point on the line, not a direction vector. Let $y=t$, then solutions are of the form $\left( \tfrac{10}{3}, t , \tfrac{2}{3}+t \right) = \left( \tfrac{10}{3}, 0 , \tfrac{2}{3} \right)+t\color{blue}{\left(0,1,1 \right)}$ and then a direction vector is $\color{blue}{(0,1,1)}$ – StackTD Apr 26 '17 at 13:27
• Yes completely forgot that I need to subtract the points to get the direction! – Yos Apr 26 '17 at 13:35
• @Yos Indeed: another way of getting a direction vector would be to subtract two points on the line. – StackTD Apr 26 '17 at 13:36

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

• I'm not familiar with the last formula in your answer, what was in the numerator and denominator? – Yos Apr 26 '17 at 13:26
• @Yos I fixed my post. See now. – Michael Rozenberg Apr 26 '17 at 13:28
• Thanks. It looks identical to the formula in my post but it's a nice example to use the $\arccos$. Is it fair to say that if I use the absolute value in the nominator I will get the value of the acute angle only? – Yos Apr 26 '17 at 13:43
• @Yos 1 $0^{\circ}\leq\measuredangle\left(L_1,L_2\right)\leq90^{\circ}$ by definition. I think your formula is wrong. – Michael Rozenberg Apr 26 '17 at 13:46

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, $$cos(\alpha)=\frac{4}{\sqrt{2}\sqrt{26}}=\frac{2\sqrt{13}}{13}$$