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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.

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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 $\theta = 90 - \alpha$.

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  • $\begingroup$ But why do you subtract ϕ from 180 to get α then do 90-α? And if you get an angle that is straightaway acute, you don't need to do 180-x first? $\endgroup$
    – user9856
    Oct 7, 2015 at 1:38
  • $\begingroup$ Note that between the normal and the line passing through the plane we have two angles, one is acute and the other is obtuse (unless the line is not in the plane, otherwise the two angles are right). When using dot product to find the angle between the line and the normal, in fact you are computing the angle between the normal vector and the directing vector of the st line. Hence, either you would get the acute angle or the obtuse angle. $\endgroup$
    – Nizar
    Oct 7, 2015 at 7:37
  • $\begingroup$ AS here you are searching for the acute one, you can get it from the result. If your result is acute so it is okay, otherwise subtract the obtained angle from 180° . $\endgroup$
    – Nizar
    Oct 7, 2015 at 7:41
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    $\begingroup$ Oh, so if the result is 90 to 180, then I got the obtuse one so 180-x gives me the acute one (then 90-theta). On the other hand, if I straightaway get an acute answer then I can assume I got the right answer then? $\endgroup$
    – user9856
    Oct 7, 2015 at 11:32
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    $\begingroup$ Exactly, this depends which angle you are searching for the acute or the obtuse . @user9856 $\endgroup$
    – Nizar
    Oct 7, 2015 at 12:24

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