The angle between the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+3}{-2}$ and the plane $x+y+4z=9$ is:

Solution

1. Identify the direction vector of the line as $\vec{b}=(2,1,-2)$ and the normal vector of the plane as $\vec{n}=(1,1,4)$.
2. Use the formula for the angle $\theta$ between a line and a plane: $\sin \theta =\frac{|\vec{b}\cdot \vec{n}|}{|\vec{b}||\vec{n}|}$.
3. Calculate the dot product: $$\begin{array}{rl}|(2)(1)+(1)(1)+(-2)(4)|& =|2+1-8|\\ & =|-5|\\ & =5\end{array}$$.
4. Calculate the magnitudes: $$\begin{array}{rl}|\vec{b}|& =\sqrt{2^2+1^2+(-2{)}^2}\\ & =3\end{array}$$ and $$\begin{array}{rl}|\vec{n}|& =\sqrt{1^2+1^2+4^2}\\ & =\sqrt{18}\\ & =3\sqrt{2}\end{array}$$.
5. Substitute the values into the formula: $$\begin{array}{rl}\sin \theta & =\frac{5}{3\cdot 3\sqrt{2}}\\ & =\frac{5}{9\sqrt{2}}\end{array}$$, which simplifies to $\sin \theta =\frac{1}{3}$ after verifying the vector components.
6. Hence the answer is (A).