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The point A, B, C, D have co-ordinates as follows $$\ A: (2,1,-2) ~~~~~~B: (4,1,-1)~~~~ ~C: (3,-2,-1)~~~~~~~ D: (3,6,2) $$

The plane R passes through A,B and C and plane Q passes through A,B and D. Find the acute angle between the two planes.

Attempt

I am aware of the fool proof method of finding the acute angle between the normal vectors of the plane. However, I tried finding the angle between AC and AD. Since one plane contains the line AC and the second contains the line AD, finding the angle between these lines should provide me the angle between the planes. However, the answer is not coming correct. Can someone point out the flaw in my method?

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  • $\begingroup$ compute the normal vectors of the given planes $\endgroup$ – Dr. Sonnhard Graubner Mar 15 '17 at 19:39
  • $\begingroup$ I am aware of that method, I want to know why the proposed shortcut is not working. $\endgroup$ – mathnoob123 Mar 15 '17 at 19:40
  • $\begingroup$ All right, you are aware of the normal vectors method, but why do you call it a "fool" method ? It's short and it is usually the only one that is usable. $\endgroup$ – Jean Marie Mar 15 '17 at 20:04
  • $\begingroup$ fool-proof. Meaning infallible method $\endgroup$ – mathnoob123 Mar 15 '17 at 20:06
  • $\begingroup$ "I want to know why the proposed shortcut is not working" -- because it's wrong? Any plane contains infinitely many lines that go in many different directions. Randomly choosing lines in the two planes can give you lots of values between these lines -- so using your logic we can get so many different "answers" for this angle. For example, you're saying that "one plane contains the line $AC$ and the second contains the line $AD$". But it's also true that one plane contains the line $AB$ and the second contains the line $AB$ -- why didn't you use them? You would get another interesting "answer". $\endgroup$ – zipirovich Mar 15 '17 at 20:09
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The definition of the dihedral angle between the planes $\alpha$ an $\beta$ is as follows: Take a plane $\gamma$ perpendicular to $\alpha$ and $\beta$. $\gamma$ will intersect $\alpha$ and $\beta$ in two lines. The angle of these two lines is the dihedral angle as shown below.

enter image description here

The short cut doesn't work because the lines $AC$ and $AD$ are not (necessarily) in a perpendicular plane.

It is an absolute theorem that the angle between lines belonging to a plane not perpendicular to $\alpha$ and $\beta$ is less than the dihedral angle.

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  • $\begingroup$ Just one more thing, if C were to revolve around A, making a horizontal circle, the angle between AD and AC will keep changing right? $\endgroup$ – mathnoob123 Mar 15 '17 at 20:13
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    $\begingroup$ @FaiqRaees: Right. $\endgroup$ – zoli Mar 15 '17 at 20:14
  • $\begingroup$ And one last thing, Is it correct to find the angle between two planes, using a line on one plane and the normal of other plane and then use the dot product to find the angle? $\endgroup$ – mathnoob123 Mar 15 '17 at 20:21
  • $\begingroup$ @FaiqRaees: If the line is not perpendicular to the intersection of the planes then no. If it is then the direction vector of that line is a normal vector to the other plane. So we are back... $\endgroup$ – zoli Mar 15 '17 at 20:34
  • $\begingroup$ Yeah that's what I was thinking. Thank you $\endgroup$ – mathnoob123 Mar 15 '17 at 20:35
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The angle between vectors in a pair of intersecting planes bears little to no relation to the angle between the planes. Consider, for example, the $x$-$y$ and $x$-$z$ planes. The angle between the planes is of course $\pi/2$, but if you take vectors nearly parallel to the $x$-axis, the angle between them is either near zero or near $\pi$.

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  • $\begingroup$ Yeah but because they are nearly parallel not completely parallel. If I were to take vector which are on the x-y plane, they will have a 90 degree angle with the x-z plane. $\endgroup$ – mathnoob123 Mar 15 '17 at 20:03
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    $\begingroup$ @FaiqRaees That’s not the point. You didn’t measure the angle between a plane and a vector in the other plane. You measured the angle between two vectors in the respective planes, which can be any value whatsoever. $\endgroup$ – amd Mar 15 '17 at 20:08
  • $\begingroup$ @FaiqRaees For that matter, the angle between a vector in the $x$-$y$ plane and the $x$-$z$ plane isn’t always 90 degrees, either. $\endgroup$ – amd Mar 15 '17 at 20:11
  • $\begingroup$ Oh okay got it. $\endgroup$ – mathnoob123 Mar 15 '17 at 20:13

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