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Find the angle between the lines whose direction cosines $l$, $m$ & $n$ are linked by the following two equations.

$$l+m+n=0$$ $$mn/(q-r)+nl/(r-p)+lm/(p-q)=0$$

Where $p$, $q$ and $r$ are constants.

Answer given is $\pi/3$

I have been trying to solve this for past hour. I have tried eliminating n from both equations and converting the second equation into a quadratic equation with variable l/m. Now the two roots of this equation will be the ratio of direction cosines of both the lines. I am unable to proceed further from there.

Formula for angle is as follows: Cos ($\theta$) = $l_1l_2+m_1m_2+n_1n_2$ where $l_1,m_1,n_1$ are direction cosines of first line and $l_2,m_2,n_2$ are direction cosines of second line

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  • $\begingroup$ Use $l^2+m^2+n^2=1$. $\endgroup$ – Jaideep Khare Dec 19 '18 at 6:56
  • $\begingroup$ Are $(p,q,r)$ the direction cosines of the second line?? $\endgroup$ – bubba Dec 19 '18 at 7:02
  • $\begingroup$ I have tried to use thus identity but I am unable to get the solution. $\endgroup$ – Sai Teja Dec 19 '18 at 7:02
  • $\begingroup$ p, q and r are constants. $\endgroup$ – Sai Teja Dec 19 '18 at 7:03
  • $\begingroup$ OK. There are two lines. What symbols denote the direction cosines of the second one? $\endgroup$ – bubba Dec 19 '18 at 7:04
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First note that $l,m,n\ne0$. This is because, say, if $l=0$, we will have $m+n=0,mn=0\therefore l^2+m^2+n^2=0\ne1$.

Now, the angle between the vectors whose direction cosines $(l_1,m_1,n_1),(l_2,m_2,n_2)$ are given by the equations is equal to the angle between the vectors $(1,m_1/l_1,n_1/l_1)\equiv(1,x_1,y_1),(1,m_2/l_2,n_2/l_2)\equiv(1,x_2,y_2)$, provided $l_1l_2>0$.

$$\displaystyle\cos\theta=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_1^2+m_1^2+n_1^2}}=\frac{l_1l_2}{|l_1l_2|}\cdot\frac{1+x_1x_2+y_1y_2}{\sqrt{1+x_1^2+y_1^2}\sqrt{1+x_2^2+y_2^2}}$$

Divide $l+m+n=0$ by $l$,$$1+x+y=0$$

Instead of juggling $3$ constants $p,q,r$, take $q-r=A,r-p=B$ and divide $\displaystyle\frac{mn}A+\frac{nl}B-\frac{lm}{A+B}=0$ by $l^2$,$$\displaystyle\frac{xy}A+\frac yB-\frac x{A+B}=0$$

Eliminate $y$ to get,

$$\displaystyle\frac{x^2}A+x\Big(\frac1A+\frac1B+\frac1{A+B}\Big)+\frac1B=0$$

Sum of roots,$$\displaystyle x_1+x_2=-A\Big[\frac1B+\frac1{A+B}\Big]-1$$

Product of roots,$$\displaystyle x_1x_2=\frac AB$$

$\displaystyle1+x_1x_2+y_1y_2=1+x_1x_2+(1+x_1)(1+x_2)=2+x_1+x_2+2x_1x_2=\frac{A^2+AB+B^2}{B(A+B)}$

$(1+x_1^2+y_1^2)(1+x_2^2+y_2^2)\\=(1+x_1^2+(1+x_1)^2)(1+x_2^2+(1+x_2)^2)\\=4(x_1^2+x_1+1)(x_2^2+x_2+1)$

Substitute for $x_1^2,x_2^2$ from the quadratic equation,

$=4\Big[1-\frac AB-Ax_1\Big(\frac1B+\frac1{A+B}\Big)\Big]\Big[1-\frac AB-Ax_2\Big(\frac1B+\frac1{A+B}\Big)\Big]\\=4\Big[\Big(1-\frac AB\Big)^2-A\Big(1-\frac AB\Big)\Big(\frac1B+\frac1{A+B}\Big)[x_1+x_2]+A^2\Big(\frac1B+\frac1{A+B}\Big)^2x_1x_2\Big]\\=\frac4{B^2(A+B)^2}\Big[(B-A)^2(B+A)^2+\frac AB(B-A)(A+2B)(A^2+3AB+B^2)+\frac{A^3}B(A+2B)^2\Big]$

We have $3$ terms in the sums. I simplified the last $2$ terms first because they have more in common.

$\displaystyle=\frac{4[A^2+B^2+AB]^2}{B^2(A+B)^2}$

We get $\displaystyle\cos\theta=\frac{1+x_1x_2+y_1y_2}{\sqrt{1+x_1^2+y_1^2}\sqrt{1+x_2^2+y_2^2}}=\pm1/2\therefore\theta=\pi/3,2\pi/3$.

In any case, since we are talking about lines which extend in both directions indefinitely and not vectors, the angle between them is often stated as the acute angle between them, given by $\pi/3$.

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  • $\begingroup$ Well done. (+1). I hope the OP appreciates all this effort. $\endgroup$ – bubba Dec 20 '18 at 1:56
  • $\begingroup$ This is exactly what I was looking for. Thank you for the answer Shubham. Really appreciate it. $\endgroup$ – Sai Teja Jan 21 '19 at 8:05
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A clarification of the question, and some steps towards a solution:

Suppose the direction cosines of the two lines are $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$. We are told (I think) that

\begin{aligned} &l_1 + m_1 + n_1 = 0 \quad ; \quad m_1n_1/(q-r)+n_1l_1/(r-p)+l_1m_1/(p-q)=0 \\ &l_2 + m_2 + n_2 = 0 \quad ; \quad m_2n_2/(q-r)+n_2l_2/(r-p)+l_2m_2/(p-q)=0 \\ \end{aligned}

where $p,q,r$ are constants. Of course, we also know that

\begin{aligned} &l_1^2 + m_1^2 + n_1^2 = 1 \\ &l_2^2 + m_2^2 + n_2^2 = 1 \end{aligned}

From these 6 equations, we need to deduce that $l_1l_2 + m_1m_2 + n_1n_2 = \tfrac12$.

I was not successful in doing this. Brute force algebra is too laborious, and there is probably some clever shortcut that I'm not able to see.

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