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.
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.
Best Answer
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).
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 $\theta = 90 - \alpha$.