Find the angle between the line \frac{{x + 1}}{2} = \frac{y}{3} = \frac{{z - 3}}{6} and the plane 10x + 2y – 11z = 3.
Find the angle between the line \frac{{x + 1}}{2} = \frac{y}{3} = \frac{{z - 3}}{6} and the plane 10x + 2y – 11z = 3.

Answer:

Direction ratios of the given line are 2,3,6

Direction ratios of the normal to the given plane are 10,2,−11

The angle between the line and the plane:

\begin{array}{l}

\sin \theta=\frac{|\vec{a} \cdot \vec{n}|}{|\vec{a}||\vec{n}|}\\

\sin \theta=\frac{\left|a_{1} \cdot a_{2}+b_{1} \cdot b_{2}+c_{1} \cdot c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2} \cdot \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\}\left\{\sqrt{\left.(10)^{2}+2^{2}+-11\right)^{2}}\right\}}\\

=\frac{40}{\{\sqrt{49}\} \times\{\sqrt{225}\}}=>\frac{40}{(7 \times 15)}=>\frac{8}{21}\\

\Rightarrow \theta=\sin ^{-1}\left(\frac{8}{21}\right)

\end{array}