How Do I Calculate the Angle Between Two Planes?

In summary: Remember that some calculators default to degrees so you may need to convert the answer.In summary, to find the angle between the planes x=0 and 2x+3y-z=4, you need to find the normal vectors of each plane and then use the dot product to calculate the angle. However, for the dot product, the normal vectors must be unit vectors, so they need to be divided by the magnitude of the normal vector. Once the dot product is calculated, the angle can be found using the inverse cosine function. If your calculator is set to degrees, you may need to convert the answer to radians.
  • #1
hkor
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how do i find the angle between the plane x=0 and the plane 2x+3y-z=4?
 
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  • #2
hkor said:
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)?
 
  • #3
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?
 
  • #4
hkor said:
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?
 
  • #5
hkor said:
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 [itex]\sqrt{4+ 9+ 1}= \sqrt{14}[/itex]
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 [itex]\sqrt{14}[/itex].

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?
 
  • #6
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?
 
  • #7
hkor said:
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.
 

What is the angle between two planes?

The angle between two planes is the measure of the angle formed by the intersection of the two planes. It is the smallest angle between two lines, one on each plane, drawn perpendicular to the intersection line.

How do you calculate the angle between two planes?

To calculate the angle between two planes, you can use the dot product formula: angle = cos⁻¹ (|a∙b| / |a| ∙ |b|), where a and b are the normal vectors of the two planes. Alternatively, you can use the cross product formula: angle = sin⁻¹ (|a x b| / |a| ∙ |b|).

Can the angle between two planes be greater than 90 degrees?

Yes, the angle between two planes can be greater than 90 degrees. It can range from 0 degrees (when the planes are parallel) to 180 degrees (when the planes are perpendicular).

What does it mean if the angle between two planes is 0 degrees?

If the angle between two planes is 0 degrees, it means that the two planes are parallel. This means that they do not intersect and have the same slope in all directions.

How can the angle between two planes be used in real life?

The concept of the angle between two planes is important in fields such as engineering, physics, and architecture. It is used to measure the orientation and relationship between two surfaces, which can be useful in designing and constructing buildings, bridges, and other structures. It is also used in navigation and 3D modeling.

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