Find the angle between the lines \(\frac{x-3}{1} = \frac{y-2}{2}=\frac {z+1}{2}\) and \(\frac{x-0}{3} = \frac{y-5}{2}=\frac {z-2}{6}\)

Concept:

The angle between the lines with direction ratios (a1,b1,c1) and (a2,b2,c2) is given by: 

\(\rm \cos {\bf{θ }} = \;\frac{{{{\bf{a}}_1}{{\bf{a}}_2} + \;{{\bf{b}}_1}{{\bf{b}}_2} + {{\bf{c}}_1}{{\bf{c}}_2}\;}}{{\sqrt {{\bf{a}}_{1\;}^2 + \;{\bf{b}}_1^2 + \;{\bf{c}}_1^2} \sqrt {{\bf{a}}_{2\;}^2 + \;{\bf{b}}_2^2 + \;{\bf{c}}_2^2} }}\)

Calculation:

Given:  \(\frac{x-3}{1} = \frac{y-2}{2}=\frac {z+1}{2}\) and \(\frac{x-0}{3} = \frac{y-5}{2}=\frac {z-2}{6}\)

As we know that, if \(\rm \frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\;and\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}\) are two lines

then the angle between them is given by: \(\rm \cos {\bf{θ }} = \;\frac{{{{\bf{a}}_1}{{\bf{a}}_2} + \;{{\bf{b}}_1}{{\bf{b}}_2} + {{\bf{c}}_1}{{\bf{c}}_2}\;}}{{\sqrt {{\bf{a}}_{1\;}^2 + \;{\bf{b}}_1^2 + \;{\bf{c}}_1^2} \sqrt {{\bf{a}}_{2\;}^2 + \;{\bf{b}}_2^2 + \;{\bf{c}}_2^2} }}\)

Here, a1 = 1, b1 = 2, c1 = 2, a2 = 3, b2 = 2 and c2 = 6

⇒ \(\rm \cos {\bf{θ }} = \;\frac{ (1)(3)+(2)(2)+(2)(6) }{{\sqrt {{1}^2 + 2^2 + 2^2} \sqrt {3^2 + 2^2 + 6^2} }}\)

\(\rm \cos {\bf{θ }} = \;\frac{19}{{\sqrt {9} \sqrt {49}}} = \frac{19}{21}\)

∴ θ = cos^-1 \(\frac{19}{21}\)