Why isn't one step enough when finding the angle between lines and planes?

Question

I wanted to ask a followup question on these questions

How to find angle between line and plane?

Finding acute angle between line and plane (Vectors)

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I want to know why 77.3 degrees isn't the correct answer. 77.3 is already acute. Why do we have to subtract 90 degrees from 77.3 and obtain 12.7 degrees as the correct answer?

In your case the actual angle should be

Step one $$ 180-102.7=77.3 $$ and as a result the angle between the plane and the line should be
Step two $$ 90-77.3=12.7 $$ instead.

Why isn't 77.3 degrees a reasonable answer for the angle between some line and a plane? I think the angle between the plane and the line isn't 12.7 degrees but rather 77.3 degrees.

As a short cut, we could have just used arcsin instead of arccosine to find the angle between the plane and the line. Why is using arcsin a good idea?

Answer 1

One nice thing about three-dimensional vectors is it's still possible to draw a reasonably intuitive graph.

The picture below is a graph of the line $(x,y,z) = t(2,4,3)$ (a line whose direction vector is $(2,4,3)$) and the plane $-5x+8y-14z = 0$ (a plane whose normal vector is $(-5,8,-14)$) drawn by https://www.geogebra.org/classic/3d.

As you can see, the angle is not very large, certainly not $77.3$ degrees. However, the normal vector $(-5,8,-14)$ (not shown here) would stick perpendicularly out from the blue plane and would make a much larger angle with the line $(x,y,z) = t(2,4,3).$

Measuring the angle between the line and the normal vector is usually much easier to do than to measure the angle between the line and the plane directly, which is why we end up going about this in such a roundabout way (calculating the angle we don't want, and then subtracting it from $90$ degrees to get the angle we do want).

Answer 2

Suppose we represent the plane with the ground. Now suppose there's a pole perpendicular to the ground with one end at the origin and the other end is at position x = -5, y = 8, z = -14. You have a stake with one end at the origin and the other at position x = 2, y = 4, z = 3. When you get that the angle between the stake and the pole is 102.7 degrees, that means that they're pointing in different directions; the pole has a z-coordinate that's negative, so it buried in the ground. If you imagine extending the pole in the other direction, it's still normal to the ground, and the angle between it and the stake is 180-102.7 = 77.3. But the angle between your stake and the ground is the complement of the angle between the angle between your stake and the pole.

If you're still confused, draw a diagram (You can simplify it down to two dimensions. The numbers for the angles will be different, but the concepts will be the same. Put your x and y axes on some graph paper. Call the origin O. Mark the point (8,-14) and call it A. That's the end of the pole. Now mark the point (-8,14), and call it B. Going from O to A is the original pole, and going from O to B should be the same line, but in the opposite direction. Now mark the point (4,3) and call it C. This is the end of your stake. Now draw the perpendicular bisector of AB. This is the ground. Pick a point D on the ground in the same direction as C. The angle AOC represents the angle you get when you use the cosine formula. The angle BOC represents the angle you get when you take the supplement of that angle. The angle DOC represents the angle your stake is to the ground. Notice that m(AOC)+m(BOC) = 180 degrees and m(BOC)+m(DOC) = 90 degrees.