Question

The direction cosines of the lines bisecting the angle between the lines whose direction cosines are ${l_1},{m_1},{n_1}$and ${l_2},{m_2},{n_2}$ and the angle between these lines is $\theta $, are
A) $\dfrac{{{l_1} - {l_2}}}{{\cos \dfrac{\theta }{2}}},\dfrac{{{m_1} - {m_2}}}{{\cos \dfrac{\theta }{2}}},\dfrac{{{n_1} - {n_2}}}{{\cos \dfrac{\theta }{2}}}$
B) $\dfrac{{{l_1} + {l_2}}}{{2\cos \dfrac{\theta }{2}}},\dfrac{{{m_1} + {m_2}}}{{2\cos \dfrac{\theta }{2}}},\dfrac{{{n_1} + {n_2}}}{{2\cos \dfrac{\theta }{2}}}$
C) $\dfrac{{{l_1} + {l_2}}}{{2\sin \dfrac{\theta }{2}}},\dfrac{{{m_1} + {m_2}}}{{2\sin \dfrac{\theta }{2}}},\dfrac{{{n_1} + {n_2}}}{{2\sin \dfrac{\theta }{2}}}$
D) $\dfrac{{{l_1} - {l_2}}}{{2\sin \dfrac{\theta }{2}}},\dfrac{{{m_1} - {m_2}}}{{2\sin \dfrac{\theta }{2}}},\dfrac{{{n_1} - {n_2}}}{{2\sin \dfrac{\theta }{2}}}$

Answer 1

Hint: Draw the diagram for the given relation in the question. Use the formula for the dot product to find the relations between the angle bisector and the given vectors. Solve these relations to find the required answer.

Complete step by step solution:
Let us assume the line with the direction cosines ${l_1},{m_1},{n_1}$ be ${v_1}$ and the line with direction cosines ${l_2},{m_2},{n_2}$ be ${v_2}$.
Let us assume the internal bisector of the lines ${v_1}$and ${v_2}$be represented by $OA$and have direction cosines as $l,m,n$, and are the required direction cosines.

The dot product of two vectors is given by $\left| A \right|\left| B \right|\cos \alpha $, where $\alpha $ is the angle between the vectors $A$ and $B$.
The dot product of the line ${v_1}$ with $OA$, we get
$\left( {{l_1}\hat i + {m_1}\hat j + {n_1}\hat k} \right)\left( {l\hat i + m\hat j + n\hat k} \right) = \sqrt {\left( {{l_1}^2 + {m_1}^2 + {n_1}^2} \right)\left( {{l^2} + {m^2} + {n^2}} \right)} \cos \dfrac{\theta }{2}$
Since we know that the sum of squares of the direction cosines of any vector is equal to 1.
Therefore, ${l_1}^2 + {m_1}^2 + {n_1}^2 = 1$and ${l^2} + {m^2} + {n^2} = 1$
Thus the equation $\left( {{l_1}\hat i + {m_1}\hat j + {n_1}\hat k} \right)\left( {l\hat i + m\hat j + n\hat k} \right) = \sqrt {\left( {{l_1}^2 + {m_1}^2 + {n_1}^2} \right)\left( {{l^2} + {m^2} + {n^2}} \right)} \cos \dfrac{\theta }{2}$ is reduced to
$\left( {{l_1}\hat i + {m_1}\hat j + {n_1}\hat k} \right)\left( {l\hat i + m\hat j + n\hat k} \right) = \cos \dfrac{\theta }{2}$
Upon simplifying, we get
$l{l_1} + m{m_1} + n{n_1} = \cos \dfrac{\theta }{2}$
Similarly, for dot product for ${v_2}$ and $OA$, we get
$l{l_2} + m{m_2} + n{n_2} = \cos \dfrac{\theta }{2}$
Adding the equations $l{l_1} + m{m_1} + n{n_1} = \cos \dfrac{\theta }{2}$ and $l{l_2} + m{m_2} + n{n_2} = \cos \dfrac{\theta }{2}$, we get
$l{l_2} + m{m_2} + n{n_2} + l{l_1} + m{m_1} + n{n_1} = \cos \dfrac{\theta }{2} + \cos \dfrac{\theta }{2}$
Upon simplifying, we get
$
  2\cos \dfrac{\theta }{2} = l\left( {{l_1} + {l_2}} \right) + m\left( {{m_1} + {m_2}} \right) + n\left( {{n_1} + {n_2}} \right) \\
  1 = \dfrac{{l\left( {{l_1} + {l_2}} \right)}}{{2\cos \dfrac{\theta }{2}}} + \dfrac{{m\left( {{m_1} + {m_2}} \right)}}{{2\cos \dfrac{\theta }{2}}} + \dfrac{{n\left( {{n_1} + {n_2}} \right)}}{{2\cos \dfrac{\theta }{2}}} \\
  \cos 0 = \dfrac{{l\left( {{l_1} + {l_2}} \right)}}{{2\cos \dfrac{\theta }{2}}} + \dfrac{{m\left( {{m_1} + {m_2}} \right)}}{{2\cos \dfrac{\theta }{2}}} + \dfrac{{n\left( {{n_1} + {n_2}} \right)}}{{2\cos \dfrac{\theta }{2}}} \\
$
The above expression is equivalent for the dot product of the vectors $OA$with direction cosines $l,m,n$ and vector with direction cosines $\dfrac{{\left( {{l_1} + {l_2}} \right)}}{{2\cos \dfrac{\theta }{2}}},\dfrac{{\left( {{m_1} + {m_2}} \right)}}{{2\cos \dfrac{\theta }{2}}},\dfrac{{\left( {{n_1} + {n_2}} \right)}}{{2\cos \dfrac{\theta }{2}}}$, with angle between them being 0.
Since the angle between the above two mentioned vectors is 0, they coincide.
Thus $\dfrac{{\left( {{l_1} + {l_2}} \right)}}{{2\cos \dfrac{\theta }{2}}},\dfrac{{\left( {{m_1} + {m_2}} \right)}}{{2\cos \dfrac{\theta }{2}}},\dfrac{{\left( {{n_1} + {n_2}} \right)}}{{2\cos \dfrac{\theta }{2}}}$ are the direction cosines of the required vector.

Therefore, option B is the correct answer.

Note: The dot product of two vectors is given by $\left| A \right|\left| B \right|\cos \alpha $, where $\alpha $ is the angle between the vectors $A$ and $B$. A visual representation of the scenario will help to formulate the equations easily. The direction cosines of any line in a three-dimensional plane denotes the cosine of the angle that is made by the line with the $x,y$ and $z$ axes. The sum of squares of the direction cosines is equal to 1.