How Do I Calculate the Angle Between Two Planes?

[hkor]

how do i find the angle between the plane x=0 and the plane 2x+3y-z=4?

 

[chiro]

how do i find the angle between the plane x=0 and the plane 2x+3y-z=4?

Hey hkor and welcome to the forums.

Do you know how to find the normal for a given plane given a plane equation?

Based on this would do you think you would do given this information to find the angle? (Hint dot product)?

 

[hkor]

Oh great thanks for the hint. I wasn't aware that for the angle between intersecting planes, the normal vector was used, unlike in the angle between intersecting lines. I think this is correct - the normal vector of 2x+3y-z=4 is (2,3,-1) and the normal vector of x=0 is (1). Thus the dot product would be (2)(1)+(3)+(-1) = 4. The magnitude would be sqrt(2^2+3^2+(-1)^2) * sqrt(1)= sqrt(14). thus the angle is equal to cos^-1 (4/sqrt(14))= cos^-1(1.069). I think this is correct however when typing the formula for the angle into my calculator a math error comes up. Can you explain this?

 

[chiro]

Oh great thanks for the hint. I wasn't aware that for the angle between intersecting planes, the normal vector was used, unlike in the angle between intersecting lines. I think this is correct - the normal vector of 2x+3y-z=4 is (2,3,-1) and the normal vector of x=0 is (1). Thus the dot product would be (2)(1)+(3)+(-1) = 4. The magnitude would be sqrt(2^2+3^2+(-1)^2) * sqrt(1)= sqrt(14). thus the angle is equal to cos^-1 (4/sqrt(14))= cos^-1(1.069). I think this is correct however when typing the formula for the angle into my calculator a math error comes up. Can you explain this?

You have to normalize the first vector for it to be "normal". Remember normal vectors in many conventions have unit one. In some textbooks normal vectors are not unit length but many applications assume they are of length one. Also remember that for |a||b|cos(theta) if |a| or |b| are not 1 then you can see that you will have problems.

So with this in mind, can you fix up your calculation?

 

[HallsofIvy]

Oh great thanks for the hint. I wasn't aware that for the angle between intersecting planes, the normal vector was used, unlike in the angle between intersecting lines. I think this is correct - the normal vector of 2x+3y-z=4 is (2,3,-1)

As chiro said, you need to use the unit normal. Divide by \sqrt{4+ 9+ 1}= \sqrt{14}

and the normal vector of x=0 is (1).

That's not even a vector! The normal vector to the yz-plane, x= 0, is (1, 0, 0), which is unit length.

Thus the dot product would be (2)(1)+(3)+(-1) = 4.

You do understand that (3) is the same as (3)(1), don't you? You are taking the dot product of (2, 3, -1) and (1, 1, 1). You want, instead, (2, 3, -1).(1, 0, 0)= 2. Remember to divide by \sqrt{14}.

magnitude would be sqrt(2^2+3^2+(-1)^2) * sqrt(1)= sqrt(14). thus the angle is equal to cos^-1 (4/sqrt(14))= cos^-1(1.069). I think this is correct however when typing the formula for the angle into my calculator a math error comes up. Can you explain this?

 

[hkor]

So when finding the dot product i use the unit vector (2/sqrt(14),3/sqrt(14),-1/sqrt(14)) . (1,0,0) = 2/sqrt(14). The magnitude of the normal vectors will be one, thus the ans will be... cos ^-1 (2/sqrt(14))= 1.0069?

 

[chiro]

So when finding the dot product i use the unit vector (2/sqrt(14),3/sqrt(14),-1/sqrt(14)) . (1,0,0) = 2/sqrt(14). The magnitude of the normal vectors will be one, thus the ans will be... cos ^-1 (2/sqrt(14))= 1.0069?

That looks correct if its in radians.