Direction ratios of the given line are 2, 3, 6
Direction ratios of the normal to the given plane are 10, 2, −11
Now, the angle between the line and the plane is given by:
\(\sin \theta = \frac{|\vec a . \vec n|}{|\vec a||\vec n|}\)
\(\sin \theta = \frac{|a_1.a_2+ b_1.b_2 + c_1.c_2|}{\sqrt{{a_1}^2 + {b_1}^2 + {c_1}^2}.\sqrt{{a_2}^2 + {b_2}^2 + {c_2}^2}}\)
\(\therefore \sin \theta = \frac{|(2\times 10) + (3\times 2) + 6 \times(-11)|}{\left(\sqrt{2^2 + 3^2 + 6^2}\right)\times\left(\sqrt{(10)^2 + 2^2 + (-11)^2}\right)}\)
\(= \frac{40}{\left(\sqrt {49}\right)\times\left(\sqrt{225}\right)}\)
\(= \frac{40}{7 \times 15}\)
\(= \frac 8 {21}\)
⇒ \(\theta = \sin^{-1} \left(\frac 8{21}\right)\)
