Formula and Method to Find the Angle Between Two Planes Using Normal Vectors
In geometry, it is important to know about various kinds of surfaces. A line is a one-dimensional surface and a space is a three-dimensional surface. However, a plane is a two dimensional surface with zero thickness. Plane is formed by some stack of lines, which are kept aside from each other. Planes play an important part of 3-D geometry. An infinite number of planes can exist in three-dimensional space and in coordinate geometry. Planes have certain special properties which include:
Any two distinct planes are either parallel or intersect at a line.
A line may either lie within the plane or intersect the plane at a single point or parallel to the plane.
Two lines are parallel to each other if they are perpendicular to the same plane.
Two planes are also parallel to each other if they are perpendicular to the same line.
Angle Between the Two Planes Formula
The angle between two planes is defined as the angle between the normal to the two planes. Two planes are referred to as perpendicular if their standard vectors are rectangular. The angle of separation of two intersecting planes is calculated as the angle of separation of normals to both planes. The angle that exists between two vectors can be determined with the help of vector multiplication.
Let us consider two planes intersecting at an angle θ as shown in the above figure. Let n1 and n2 be the normal vectors drawn to the planes. The equation for both the planes is thus given as
\[\bar{r}\].\[\bar{n_{1}}\] = \[d_{1}\]
\[\bar{r}\].\[\bar{n_{2}}\] = \[d_{2}\]
Cosine of angle between two intersecting planes is given as the cosine of the angle between their normals:
This is the angle between two planes formula when normal vectors are given.
Angle between Two Planes Formula in Cartesian System
Consider the angle between two planes, for example in which two planes intersect at an angle θ. In the cartesian system points are labelled on a plane. A plane is formed with two perpendicular lines called the X-axis and the Y-axis, this is called the cartesian or coordinate plane.
The equations of these two planes are given in the cartesian coordinate system as A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0. In these two equations, A1, B1 and C1 are the direction ratios of normal to the plane described by the equation A1x + B1y + C1z + D1 = 0 and A2, B2 and C2 are the direction ratios of normal to the plane defined by the equation A2x + B2y + C2z + D2 = 0.
Now consider the angle between the normal to the two planes and (A1, B1, C1) and (A2, B2, C2) are the direction ratios of the normal to both the planes in consideration. The cosine of the angle between the two planes is given as:
Solved Examples of Angle Between Two Planes
1. Find the angle between two planes r.(2i - j + k) = 1 and r.(i + k) = -3
Solution: For finding the angle between the two planes r.(2i - j + k) = 1 and r.(i + k) = -3, Using the formula Cosθ = \[\frac{n_{1}.n_{2}}{|n_{1}||n_{2}|}\].
n1 = 2i - j + k, n2 = i + k
\[ |n_{1}| = \sqrt{(2^{2} + (-1)^{2} + 1^{2})} = \sqrt{(4 + 1 + 1)} = \sqrt{6} \]
\[ |n_{2}| = \sqrt{(1^{2} + (0)^{2} + 1^{2})} = \sqrt{(1 + 0 + 1)} = \sqrt{2} \]
Scalar product of the normal vectors is,
n1 . n2 = (2i - j + k) . (i + k) = 2 × 1 + (-1) × (0) + 1 × 1 = 2 + 0 + 1 = 3
Substituting the values into the formula,
\[ cos θ = \frac{|(3)|}{(\sqrt{6} . \sqrt{2}})\]
= \[\frac{3}{\sqrt{12}}\]
= \[\frac{\sqrt{3}}{2}\]
\[ ⇒ θ = cos^{-1}(\frac{\sqrt{3}}{2}) \]
= π/6 radians
Therefore the angle between two planes r.(2i - j + k) = 1 and r.(i + k) = -3 is equal to π/6 radians.
2. Determine the angle between two planes P1: 3x - 6y + 2z = 7 and P2: 2x + 2y - 2z = 3
Solution: For determining the angle between two planes in cartesian form using the formula cos θ = \[\frac{A_{1}A_{2}+B_{1}B_{2}+C_{1}C_{2}}{\sqrt{A_{1}^{2}+B_{1}^{2}+C_{1}^{2}}.\sqrt{A_{2}^{2}+B_{2}^{2}+C_{2}^{2}}}\].
The equations of the planes are P1: 3x - 6y + 2z = 7 and P2: 2x + 2y - 2z = 3.
Now, A1 = 3, B1 = -6, C1 = 2, A2 = 2, B2 = 2, C2= -2.
Substituting the values into the formula,
\[ cos θ = \frac{|(3\times 2 + (-6)\times 2 + 2\times (-2))|}{\sqrt{(32 + (-6)2 + 22)} \sqrt{(22 + 22 + (-2)2)}} \]
= \[ \frac{|(6 + (-12) - 4)|}{\sqrt{(9 + 36 + 4)}\sqrt{(4 + 4 + 4)}} \]
= \[ \frac{|-10|}{(\sqrt{49} \sqrt{12})} \]
= \[ \frac{10}{(7\times 2\sqrt{3})} \]
= \[ \frac{5}{7\sqrt{3}} \]
= \[ \frac{5\sqrt{3}}{21} \]
\[ θ = cos^{-1}(\frac{5\sqrt{3}}{21}) \]
Therefore the angle between the two planes P1: 3x - 6y + 2z = 7 and P2: 2x + 2y - 2z = 3 is \[ cos^{-1}(\frac{5\sqrt{3}}{21}) \].
3. How to calculate angle between two planes described by the equations 2x + 4y - 4z - 6 = 0 and 4x + 3y + 9 = 0?
Solution: In these equations of the plane,
A1 = 2, B1 = 4, C1 = - 4, D1 = - 6
A2 = 4, B2 = 3, C2 = 0, D2 = 9
Cosθ = \[\frac{A_{1}A_{2}+B_{1}B_{2}+C_{1}C_{2}}{\sqrt{A_{1}^{2}+B_{1}^{2}+C_{1}^{2}}.\sqrt{A_{2}^{2}+B_{2}^{2}+C_{2}^{2}}}\]
Cosθ = \[\frac{|2 \times 4+4 \times 3+(-4) \times 0|}{\sqrt{2^{2}+4^{2}+(-4)^{2}}.\sqrt{4^{2}+3^{2}+0^{2}}}\]
Cosθ = \[\frac{|8+12+0|}{\sqrt{4+16+16}.\sqrt{16+9+0}}\]
Cosθ = \[\frac{20}{\sqrt{36}.\sqrt{25}}\] = \[\frac{20}{6 \times 5}\] = \[\frac{2}{3}\]
θ = \[Cos^{-1}(\frac{2}{3})\]
4. How to calculate angle between two planes when the direction vectors of normals of the planes are given as n1 = 2i + 4j - 2k and n2 = 6i - 8j - 2k.
Solution:
The coordinates of normal vector n1 is (2, 4, -2)
The coordinates of normal vector n2 is (6, -8, -2)
Cosθ = \[\frac{n_{1}.n_{2}}{|n_{1}||n-{2}|}\]
Cosθ = \[\frac{(2,4,-2)(6,-8,-2)}{|n1||n2|}\]
\[\frac{2\sqrt{39}}{\sqrt{4+16+4}.\sqrt{36+64+4}}\] = \[\frac{2\sqrt{39}}{39}\] = \[\frac{2}{\sqrt{39}}\]
θ = \[Cos^{-1}\frac{2}{\sqrt{39}}\]
Fun Facts
Point is a dimensionless geometric shape.
A plane is formed by a stack of lines arranged side by side.
Conclusion
In this article, we have explored the concept of the angle between two planes in the cartesian system. We have explained a few examples above based on these formulas for your better understanding of the concept.
The angle between two given planes can be determined by the angle present in between the normals of the two planes. The angle that lies between the two planes is also known as the dihedral angle. In other words, the dihedral angle between two planes is the angle between the two intersecting planes.
FAQs on Angle Between Two Planes in Three Dimensional Geometry
1. What is the angle between two planes?
The angle between two planes is the angle between their normal vectors at the line of intersection. In three-dimensional geometry, when two planes intersect, they form a dihedral angle. This angle is measured by finding the angle between the normal vectors of the planes using the dot product formula. The angle is always taken as the acute angle (between 0° and 90°).
2. What is the formula for the angle between two planes?
The formula for the angle between two planes is cosθ = |(n₁ · n₂)| / (|n₁||n₂|), where n₁ and n₂ are their normal vectors. Here:
- n₁ · n₂ is the dot product of the normals
- |n₁| and |n₂| are their magnitudes
- θ is the angle between the planes
3. How do you find the angle between two planes step by step?
To find the angle between two planes, calculate the angle between their normal vectors using the dot product formula. Follow these steps:
- Write each plane in the form ax + by + cz + d = 0
- Identify the normal vectors n₁ = (a₁, b₁, c₁) and n₂ = (a₂, b₂, c₂)
- Compute n₁ · n₂
- Find |n₁| and |n₂|
- Use cosθ = |(n₁ · n₂)| / (|n₁||n₂|)
- Find θ using the inverse cosine
4. Can you give an example of finding the angle between two planes?
Yes, for planes 2x + y − z = 0 and x − y + 2z = 0, the angle between them is found using their normal vectors.
- n₁ = (2, 1, −1)
- n₂ = (1, −1, 2)
- n₁ · n₂ = (2)(1) + (1)(−1) + (−1)(2) = −1
- |n₁| = √6, |n₂| = √6
- cosθ = |−1| / (√6 × √6) = 1/6
5. Why is the angle between two planes equal to the angle between their normal vectors?
The angle between two planes equals the angle between their normal vectors because normals are perpendicular to the planes. Since each normal represents the orientation of its plane, the angle between the normals directly measures how much the planes tilt relative to each other. Therefore, calculating the angle between normal vectors gives the correct dihedral angle between planes.
6. What is the angle between two parallel planes?
The angle between two parallel planes is 0°. Parallel planes have proportional normal vectors, meaning their normals point in the same or opposite direction. Since the angle between proportional vectors is 0° (or 180°, but we take the acute angle), the angle between parallel planes is always 0°.
7. What is the angle between two perpendicular planes?
The angle between two perpendicular planes is 90°. Two planes are perpendicular when their normal vectors satisfy n₁ · n₂ = 0. A zero dot product means the normals are orthogonal, which implies the planes intersect at a right angle.
8. How do you know if two planes are parallel or perpendicular?
Two planes are parallel if their normal vectors are proportional and perpendicular if their dot product is zero.
- Parallel planes: n₁ = k n₂ for some constant k
- Perpendicular planes: n₁ · n₂ = 0
9. Is the angle between two planes always acute?
Yes, the angle between two planes is always taken as the acute angle between 0° and 90°. Even if the calculated angle between normal vectors is obtuse, we use the absolute value in cosθ = |(n₁ · n₂)| / (|n₁||n₂|) to ensure the smaller angle is obtained. This convention defines the standard dihedral angle between planes.
10. What are common mistakes when finding the angle between two planes?
Common mistakes when finding the angle between two planes include using direction vectors instead of normal vectors and forgetting the absolute value in the formula. Key errors to avoid:
- Not writing the plane in standard form ax + by + cz + d = 0
- Using incorrect normal vectors
- Forgetting magnitudes in the denominator
- Not taking the absolute value in the cosine formula
