Question
What is the value of $$n$$ so that the angle between the lines having direction ratios $$\left( {1,\,1,\,1} \right)$$ and $$\left( {1,\, - 1,\,n} \right)$$ is $${60^ \circ }\,?$$
A.
$$\sqrt 3 $$
B.
$$\sqrt 6 $$
C.
$$3$$
D.
None of these
Answer :
$$\sqrt 6 $$
Solution :
If $$\left( {{l_1},\,{m_1},\,{n_1}} \right)$$ and $$\left( {{l_2},\,{m_2},\,{n_2}} \right)$$ are the direction ratios then angle between the lines is
$$\eqalign{
& \cos \,q = \frac{{{l_1}{l_2} + {m_1}{m_2} + \,{n_1}{n_2}}}{{\sqrt {l_1^2 + m_1^2 + n_1^2} \,\sqrt {l_2^2 + m_2^2 + n_2^2} }} \cr
& {\text{Here, }}{l_1} = 1,\,{m_1} = 1,\,{n_1} = 1\,{\text{and}} \cr
& {l_2} = 1,\,{m_2} = - 1,\,{n_2} = n{\text{ and }}q = {60^ \circ } \cr
& \therefore \,\cos \,{60^ \circ } = \frac{{1 \times 1 + 1 \times \left( { - 1} \right) + 1 \times n}}{{\sqrt {{1^2} + {1^2} + {1^2}} \times \sqrt {{1^2} + {1^2} + {n^2}} }} \cr
& \Rightarrow \frac{1}{2} = \frac{n}{{\sqrt 3 \sqrt {2 + {n^2}} }} \cr
& \Rightarrow {n^2} = 6 \cr
& \Rightarrow n = \pm \sqrt 6 \cr} $$