JEE MAIN - Mathematics (2014 (Offline) - No. 20)
The angle between the lines whose direction cosines satisfy the equations $$l+m+n=0$$ and $${l^2} = {m^2} + {n^2}$$ is :
$${\pi \over 6}$$
$${\pi \over 2}$$
$${\pi \over 3}$$
$${\pi \over 4}$$
Explanation
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 $$n=0$$
$$\left( {{l_2},{m_2},{n_2}} \right) = \left( { - {1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }},0} \right)$$
$$\therefore$$ $$\cos \theta = {1 \over 2} \Rightarrow \theta = {\pi \over 3}$$
$$l + m + n = 0$$ 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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