Question
What is the angle between the lines $$\frac{{x - 2}}{1} = \frac{{y + 1}}{{ - 2}} = \frac{{z + 2}}{1}$$ and $$\frac{{x - 1}}{1} = \frac{{2y + 3}}{3} = \frac{{z + 5}}{2}\,?$$
A.
$$\frac{\pi }{2}$$
B.
$$\frac{\pi }{3}$$
C.
$$\frac{\pi }{6}$$
D.
None of the above
Answer :
$$\frac{\pi }{2}$$
Solution :
$$\eqalign{
& {\text{The given lines are : - }} \cr
& \frac{{x - 2}}{1} = \frac{{y - \left( { - 1} \right)}}{{ - 2}} = \frac{{z - \left( { - 2} \right)}}{1}{\text{ and}} \cr
& \frac{{x - 1}}{1} = \frac{{y - \left( { - \frac{3}{2}} \right)}}{{\frac{3}{2}}} = \frac{{z - \left( { - 5} \right)}}{2}\, \cr
& {\text{dr's of Ist line are}}\,:{\text{ - }} \cr
& {a_1} = 1,\,{b_1} = - 2,\,{c_1} = 1 \cr
& {\text{dr's of IInd line are}}\,{\text{: - }} \cr
& {a_2} = 2,\,{b_2} = 3,\,{c_2} = 4 \cr
& {\text{Let }}'\theta '\,{\text{be the angle between two lines, then,}} \cr
& \cos \,\theta = \frac{{\left| {{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \right|}}{{\sqrt {a_1^2 + b_1^2 + c_1^2} .\sqrt {a_2^2 + b_2^2 + c_2^2} }} \cr
& \Rightarrow \cos \,\theta = 0 \cr
& \Rightarrow \theta = \frac{\pi }{2} \cr} $$