The angle between the lines whose direction cosines satisfy the equations l+m+n=0 and l^2=…

Question

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2=m^2+n^2$ is

Accepted Answer

Given

$l + m + n = 0$ and ${l^2} = {m^2} + {n^2}$

Now, $${\left( { - m - n} \right)^2} = {m^2} + {n^2}$$

$$ \Rightarrow mn = 0 \Rightarrow m = 0\,\,$$ or $$\,\,n = 0$$

If $m=0$ then $l=-n$

We know

$${l^2} + {m^2} + {n^2} = 1 \Rightarrow n =  \pm {1 \over {\sqrt 2 }}$$

i.e.$$\left( {{l_1},{m_1},{n_1}} \right) = \left( { - {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }}} \right)$$

If $n=0$ then $l=-m$

$${l^2} + {m^2} + {n^2} = 1\,\,\, \Rightarrow 2{m^2} = 1$$

$$ \Rightarrow m =  \pm {1 \over {\sqrt 2 }}$$

Let $$m = {1 \over {\sqrt 2 }} \Rightarrow l =  - {1 \over {\sqrt 2 }}$$

and $n=0$

$$\left( {{l_2},{m_2},{n_2}} \right) = \left( { - {1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }},0} \right)$$

$$\therefore$$ $$\cos \theta  = {1 \over 2} \Rightarrow \theta  = {\pi  \over 3}$$

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