The plane $2x-3y+6z-11=0$ makes an angle $sin^{-1}\alpha $ with x-axis. The value of $\alpha $ is equal to :
Answer & explanation
Correct answer: option 2
The correct answer is option (2) → $\frac{2}{7}$
$\vec n=2\hat i-3\hat j+6\hat k$
plane makes $\sin^{-1}α$ with x axis
⇒ Normal makes $(90°-\sin^{-1}α)$ with x axis.
Let x axis be $\hat b=\hat i$
so $\vec n.\hat b=|\vec n||\hat b|\cos(90°-\sin^{-1}α)$
so $2=7\sin(\sin^{-1}α)⇒α=\frac{2}{7}$