# Condition when angle between two lines is ${\pi\over 3}$

The whole question is-

Show that if the angle between the lines whose direction cosines are given $$l+m+n=0$$ and $$fmn+gnl+hlm=0$$ is $${\pi\over 3}$$ then $${1\over f}+{1\over g}+{1\over h}=0$$.

I'm trying to solve the problem in the following manner-
From the first equation $$n=-l-m$$, substituting this value of $$n$$ in the second equation we get-
$$fm(-l-m)+g(=l-m)l+hlm=0$$
$$\implies g\left({l\over m}\right)^2+(f+g+h)\left({l\over m}\right)+f=0$$
Now, the roots of this equation are $${l_1\over m_1}$$ and $${l_2\over m_2}$$. So, product of them $${l_1\over m_1}.{l_2\over m_2}={f\over g}\implies {l_1 l_2\over f}={m_1 m_2\over g}$$.
Similarly, we get $${m_1 m_2\over g}={n_1 n_2\over h}$$.
Hence, $${l_1 l_2\over f}={m_1 m_2\over g}={n_1 n_2\over h}=K$$(say)
Thus, $$\cos {\pi\over3}=l_1 l_2+m_1 m_2+n_1n_2=K(f+g+h)\implies K=\frac{\sqrt{3}}{2(f+g+h)}$$.
Now, I can't proceed further. I can't prove $${1\over f}+{1\over g}+{1\over h}=0$$.
Can anybody solve the problem? Thanks for assistance in advance.

• The direction cosines (in $3D$) are (three) numbers, not an equation, aren't they? – user Feb 28 '19 at 17:47
• Yes, I am saying that the direction cosines of the two lines $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$ satisfy these equations. – MathBS Mar 2 '19 at 19:00
• This is still not clear. Do you mean that the direction cosines are $(l,m,n)$ and $(fmn,gnl,hlm)$, respectively? – user Mar 2 '19 at 20:03
• If $(l_1,m_1,n_1), (l_2,m_2,n_2)$ are the direction cosines of two lines respectively. Then $l_1+m_1+n_1=0, fm_1 n_1+gn_1 l_1+hl_1 m_1=0$ and $l_2+m_2+n_2=0, fm_2 n_2+gn_2 l_2+hl_2 m_2=0$ – MathBS Mar 4 '19 at 15:17

As clarified in the comments, the correct formulation of the problem is:

Show that the angle between the lines whose direction cosines $$(l_1,m_1,n_1)$$ and $$(l_2,m_2,n_2)$$ satisfy the equations $$l+m+n=0\tag1$$ and $$fmn+gnl+hlm=0\tag2$$ is $${\pi\over 3}$$ if $${1\over f}+{1\over g}+{1\over h}=0.\tag3$$

In fact as will be shown below even stronger statement with "if" replaced by "if and only if" holds. It will be additionally assumed $$(l_1,m_1,n_1)\nparallel(l_2,m_2,n_2)$$, $$(f,g,h)\ne(0,0,0)$$. Otherwise extra solutions are possible.

The entirety of the conditions implies that neither of $$l_i,m_i,n_i$$ is $$0$$. Indeed an assumption that any of the components - say $$n_1$$ - is $$0$$ results in combination with $$(1)$$ and $$(2)$$ in the equality $$(f,g,h)=(0,0,0)$$.

Observe now that equality $$(1)$$ $$(l,m,n)\cdot(1,1,1)=0$$ means that the vectors $${\bf a}_1=(l_1,m_1,n_1)$$ and $${\bf a}_2=(l_2,m_2,n_2)$$ are orthogonal to the vector $${\bf 1}=(1,1,1)$$. This implies $${\bf a}_1\times {\bf a}_2 \parallel {\bf 1}$$ or component-wise: $$m_1n_2-n_1m_2=n_1l_2-l_1n_2=l_1m_2-m_1l_2=\frac{\sin\alpha}{\sqrt3},\tag4$$ where $$\alpha$$ is the angle between $${\bf a}_1$$ and $${\bf a}_2$$ which is to be found. The last equality is due to $$\sin^2\alpha=|{\bf a}_1\times {\bf a}_2|^2=\sum_{i=x,y,z}({\bf a}_1\times {\bf a}_2)_i^2.$$

Similarly the equation $$(2)$$ $$(f,g,h)\cdot(mn,nl,lm)=0$$ means that the vectors $${\bf b}_1=(m_1n_1,n_1l_1,l_1m_1)$$ and $${\bf b}_2=(m_2n_2,n_2l_2,l_2m_2)$$ are orthogonal to the vector $${\bf c}=(f,g,h)$$. This in turn implies that the vector $${\bf c}$$ is collinear to the vector product $${\bf b}_1\times {\bf b}_2$$, which reads component-wise: \begin{align} (f,g,h)& =A\big(l_1l_2(n_1m_2-m_1n_2),m_1m_2(l_1n_2-n_1l_2),n_1n_2(m_1l_2-l_1m_2)\big)\\ &=\frac{A\sin\alpha}{\sqrt3}(l_1l_2,m_1m_2,n_1n_2)\tag5 \end{align} where $$A$$ is a non-zero constant.

We start now with the proof of the "if" part. The equations $$(3)$$ and $$(5)$$ imply: $$\frac{1}{l_1l_2}+\frac{1}{m_1m_2}+\frac{1}{n_1n_2}=0\tag6$$ or $$m_1m_2n_1n_2+n_1n_2l_1l_2+l_1l_2m_1m_2=0.\tag7$$ With help of $$(1)$$ the equation reads: \begin{align} 0&=(l_1+m_1)(l_2+m_2)(l_1l_2+m_1m_2)+l_1l_2m_1m_2\\ &=(l_1l_2+m_1m_2+l_1m_2)(l_1l_2+m_1m_2+m_1l_2).\tag8 \end{align} Now we can proceed with computation of $$\cos\alpha={\bf a}_1\cdot {\bf a}_2$$: \begin{align} \cos\alpha &=l_1l_2+m_1m_2+n_1n_2\\ &=l_1l_2+m_1m_2+(l_1+m_1)(l_2+m_2)\\ &=2(l_1l_2+m_1m_2)+l_1m_2+m_1l_2\\ &\stackrel{(8)}=\pm(l_1m_2-m_1l_2)\stackrel{(4)}=\pm\frac{\sin\alpha}{\sqrt3}.\tag9 \end{align} Squaring the equation one finally obtains: $$\cos^2\alpha=\frac13 \sin^2\alpha\implies \cos^2\alpha=\frac14 \implies \alpha=\frac\pi3.\tag{10}$$

As already mentioned the inverse implication $$\alpha={\pi\over 3}\implies {1\over f}+{1\over g}+{1\over h}=0$$ holds as well, since, provided the equality $$(1)$$, the equations $$(6)-(10)$$ are equivalence relations, so that the whole argument can be easily reversed.

In general if a, b, c and d, e, f are components of 2 vectors then the direction cosines are for the first vector : $$\frac{a}{\sqrt{a^2 + b^2 + c^2}}$$ and $$\frac{b}{\sqrt{a^2 + b^2 + c^2}}$$ and $$\frac{c}{\sqrt{a^2 + b^2 + c^2}}$$ similarly for the second vector.

Then the angle $$\theta$$ between these vectors is

$$cos\theta = \frac{ad + be + cf}{\sqrt{a^2 + b^2 + c^2}\sqrt{d^2 + e^2 + f^2}}$$

Using the definition of dot product. In other words $$cos\theta$$ equals the sum of product of the directional cosines of corresponding vectors. In this question we have

$$cos\theta = l(fmn) + m(gnl) + n(hlm) = lmn(f + g + h)$$

The next step is to prove this equal 1/2.

From the question we have

$$l + m + n = 0$$ $$l^2 + m^2 + n^2 = 1$$ Since there are three unknowns but only 2 equations, we can only express l and n in terms of m :

$$\implies l^2 + lm + m^2 = \frac{1}{2}$$

$$\implies l = \frac{-m + D}{2}$$ $$\implies n = \frac{-m - D}{2}$$ Where $$D = \sqrt{2 - 3m^2}$$. It does not matter which root we take for l the other root will be for n.

Also we have other conditions:

$$fmn + gnl + hlm = 0 ... (1)$$ $$(fmn)^2 + (gnl)^2 + (hlm)^2 = 1 ... (2)$$ $$\frac{1}{f} + \frac{1}{g} + \frac{1}{h} = 0 \implies fg + gh + fh = 0... (3)$$

Put l and n into (1) we have $$-fm(m + D) - g(1 - 2m^2) + hm(-m + D) = 0...(4)$$

Put l and n into (2) we have $$f^2m^2(2 - 2m^2 + 2mD) + g^2(1 - 2m^2)^2 + h^2m^2(2 - 2m^2 -2mD) = 4...(5)$$

Eliminate f from (3) and (4) we have $$f^2m(m + D) + fg(2mD - 2m^2 + 1) + g^2(1 - 2m^2) = 0$$ $$\implies f = \frac{2m^2 - 2mD - 1\pm\sqrt{(2mD -2m^2 + 1)^2 - 4(m + D)m(1 - 2m^2)}}{2m(m + D)}g$$ $$\implies f = \frac{2m^2 - 2mD - 1 \pm 1}{2m(m + D)}g$$ We take plus 1 for simplicity since the other root may give a result making $$cos\theta$$ greater than 1 or less than -1. $$f = \frac{m - D}{m + D}g$$

Eliminate h from (4) and (5) gives $$f^2m^2(2 - 2m^2 + 2mD) + g^2(1 - 2m^2)^2 + fgm(m + D)(1 - 2m^2) = 2$$ Put f into this equation we get $$m^2(m - D)^2g^2 + g^2(1 - 2m^2)^2 + (m - D)mg^2(1 - 2m^2) = 2$$ $$\implies g^2(1 - Dm - m^2) = 2$$ $$\implies g = \sqrt{\frac{2}{1 - Dm - m^2}}$$ $$\implies f = \frac{m - D}{m + D}\sqrt{\frac{2}{1 - Dm - m^2}}$$ $$\implies h = \frac{D - m}{2m}\sqrt{\frac{2}{1 - Dm - m^2}}$$

Hence put l, m, n, f, g, h to the cosine expression we get $$cos\theta = lmn(f + g +h) = \frac{-(m - D)}{2}m\frac{-(m + D)}{2}\sqrt{\frac{2}{1 - Dm - m^2}}(\frac{m - D}{m + D} + 1 + \frac{D - m}{2m})$$ $$=\frac{m - D}{4}\sqrt{\frac{2}{1 - Dm - m^2}}$$ $$=\frac{1}{4}\sqrt{\frac{2(m^2 + 2 - 3m^2 - 2Dm)}{1 - Dm - m^2}} = \frac{1}{2}$$