Finding acute angle between line and plane (Vectors) This is a very very simple question about vectors: To find the **acute** angle between a line and a plane, you use the formula **cosx = (scalar product between normal of plane and directional vector of line)/(product of modulus of normal and directional vector)** After that, do **90 degrees minus x** to get the answer. (By the way, I realise that using sinx=.... would spare me the effort of minusing - but is that way recommended?) In this question of plane 2x-y+4z=9 and directional vector (10,5,-5), after doing all the working, I get cosx= -0.089... and x=95.11 degrees. ![enter image description here]( But the question wants **acute** angle and 90-x gives me -5.1 degrees. What should I do? The answer is 5.1 deg but I don't see how 90-x gets me that... NB: using sinx gives me -5.1 degrees straightaway. Thank you.

Honestly, between the direction vector $ v=(10,5,-5)$ and the normal vector $( 2,-1,4)$ the angle is not acute, in fact it is obtuse (you can see that by a simple plot). ![enter image description here](

However as you are asking about the angle between a line and a plane, so the you must take care of the orientation of the vectors you are working with.

In you case, to find the angle $ \theta $ you can do the following : when finding $\cos x$, apply $\arccos$ to find an angle $\phi$. Then subtract $\phi $ from $180$ to get $ \alpha=180 - \phi$ . Now $\theta = 90 - \alpha$.