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Question

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


Solution

Angle between two lines is:
$$\begin{array}{l} \cos  \theta =\dfrac { { { a_{ 1 } }{ a_{ 2 } }+{ b_{ 1 } }{ b_{ 2 } }+{ c_{ 1 } }{ c_{ 2 } } } }{ { \sqrt { { a_{ 1 } }^{ 2 }+{ b_{ 1 } }^{ 2 }+{ c_{ 1 } }^{ 2 } } \sqrt { { a_{ 2 } }^{ 2 }+{ b_{ 2 } }^{ 2 }+{ c_{ 2 } }^{ 2 } }  } }  \\ l+m+n=0.......(1) \\ l=-\left( { m+n } \right)  \\ { \left( { m+n } \right) ^{ 2 } }={ l^{ 2 } }........(2) \\ { m^{ 2 } }+{ n^{ 2 } }+2mn={ m^{ 2 } }+{ n^{ 2 } } \\ \left[ { { l^{ 2 } }={ m^{ 2 } }+{ n^{ 2 } },\, given } \right]  \\ 2mn=0 \\ When\, m=0\Rightarrow l=-n \\ Hence,\left( { l,m,n } \right) \, is\, \left( { 1,0,-1 } \right)  \\ When\, n=0,\, then\, l=-m \\ Hence,\, \left( { l,m,n } \right) \, is\, \left( { 1,0,-1 } \right)  \\ \cos  \theta =\dfrac { { 1+0+0 } }{ { \sqrt { 2 } \sqrt { 2 }  } } =\dfrac { 1 }{ 2 } \Rightarrow \theta =\dfrac { \pi  }{ 3 }  \end{array}$$

Mathematics

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