Find the angle between the line \(\frac{{x - 2}}{1} = \frac{{y + 3}}{{ - \;2}} = \frac{{z + 4}}{- \ 3}\) and the plane 2x - 3y + z - 5 = 0 ?

Concept:

If θ is the angle between the line \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) and the plane a₂x + b₂ y + c₂z + d = 0 then

\(\sin \theta = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right)\left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\)

Calculation:

Given: Equation of line is \(\frac{{x - 2}}{1} = \frac{{y + 3}}{{ - \;2}} = \frac{{z + 4}}{- \ 3}\) and equation of plane is 2x - 3y + z - 5 = 0.

As we know that the angle between the line \(\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}\) and the plane a2x + b2 y + c2z + d = 0  is given by: \(\sin \theta = \frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\left( {\sqrt {a_1^2 + b_1^2 + c_1^2} } \right)\left( {\sqrt {a_2^2 + b_2^2 + c_2^2} } \right)}}\)

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

⇒ a₁⋅ a₂ + b₁ ⋅ b₂ + c₁ ⋅ c₂ = 2 + 6 - 3 = 5

\(⇒ \sqrt {a_1^2 + b_1^2 + c_1^2} = \sqrt {14} \) and \(\;\sqrt {a_2^2 + b_2^2 + c_2^2} = \sqrt {14} \)

\(\Rightarrow \sin \theta = \frac{5}{{14}}\)

\(\Rightarrow \theta = {\sin ^{ - 1}}\left( {\frac{5}{{14}}} \right)\)