Angle Between Two Planes
The angle between two planes is determined by the angle between the normals of the two planes. It can be determined using the vector form and cartesian form equation of the plane. The angle between two planes in vector form can be determined using the dot product of the normal vectors to the two planes. We can determine the angle between two planes in vector form and cartesian form and is equal to the angle between the normal to the two planes.
The angle between the two planes is also called the dihedral angle. In this article, we will explore the concept of the angle between two planes and its formula in vector form and cartesian form. We will solve a few examples based on these formulas for a better understanding of the concept.
1.  What is Angle Between Two Planes? 
2.  Angle Between Two Planes Formula 
3.  Angle Between Two Planes in Vector Form 
4.  Angle Between Two Planes in Cartesian Form 
5.  FAQs on Angle Between Two Planes 
What is Angle Between Two Planes?
The angle between two planes is equal to the angle between the normal vectors to the two planes. We can determine the angle between two planes using the cartesian equation of the plane and the vector equation of the plane. Since the angle between the two planes is given by the angle between the normals to these two planes, therefore we use the scalar product and magnitudes of the normal vectors in the formula to find the angle between them. In vector form, the equation of a plane is given by r.n = d, and its cartesian equation is given by, Ax + By + Cz + D = 0. Now, let us go through the formulas to find the angle between two planes.
Angle Between Two Planes Formula
Now, there are two formulas to find the angle between two planes. The formulas exist in vector form and cartesian form. Consider two planes P_{1} and P_{2} and the angle between them is θ. The equations of the two planes in vector form are r.n_{1} = d_{1} and r.n_{2} = d_{2} and the equations of the two planes in the cartesian form are A_{1}x + B_{1}y + C_{1}z + D_{1} = 0 and A_{2}x + B_{2}y + C_{2}z + D_{2} = 0. Then, the formulas to find the angle between two planes are:
 cos θ = (n_{1} . n_{2})/(n_{1}.n_{2})
 cos θ = (A_{1}A_{2} + B_{1}B_{2} + C_{1}C_{2})/[√(A_{1}^{2} + B_{1}^{2} + C_{1}^{2})√(A_{2}^{2} + B_{2}^{2} + C_{2}^{2})]
Using the above formulas, we can determine the value of cos θ and take cos inverse on both sides to find the value of θ, and hence, the angle between two planes.
Angle Between Two Planes in Vector Form
Let us now solve an example based on the formula of the angle between two planes in vector form. For planes, r.n_{1} = d_{1} and r.n_{2} = d_{2}, we will use the formula cos θ = (n_{1} . n_{2})/(n_{1}.n_{2}), where n_{1} , and n_{2} are normal vectors to the two planes and θ is the angle between the two planes.
Example: Determine the angle between the two planes whose vector equations are given as r.(2i + 2j  3k) = 4 and r.(3i  3j + 5k) = 3.
Solution: The equations of the planes are given in vector form. Now to find the angle between the planes r.(2i + 2j  3k) = 4 and r.(3i  3j + 5k) = 3, we will use the formula cos θ = (n_{1} . n_{2})/(n_{1}.n_{2}). We have,
n_{1} = 2i + 2j  3k, n_{2} = 3i  3j + 5k
n_{1} = √(2^{2} + 2^{2} + (3)^{2}) = √(4 + 4 + 9) = √17
n_{2} = √(3^{2} + (3)^{2} + 5^{2}) = √(9 + 9 + 15) = √43
Scalar product of the normal vectors is given by, n_{1} . n_{2} = (2i + 2j  3k) . (3i  3j + 5k) = 2 × 3 + 2 × (3) + (3) × 5 = 6  6  15 = 15
Substituting these values into the formula, we have
cos θ = (15)/(√17 . √43)
= 15/√731
⇒ θ = cos^{1}(15/√731) [Taking inverse cosine on both sides]
= 0.983
Hence, the angle between two planes r.(2i + 2j  3k) = 4 and r.(3i  3j + 5k) = 3 is equal to cos^{1}(15/√731) = 0.983 radians.
Angle Between Two Planes in Cartesian Form
In this section, we will solve an example and find the angle between two planes with equations in cartesian form. For planes, A_{1}x + B_{1}y + C_{1}z + D_{1} = 0 and A_{2}x + B_{2}y + C_{2}z + D_{2} = 0, we will use the formula cos θ = (A_{1}A_{2} + B_{1}B_{2} + C_{1}C_{2})/[√(A_{1}^{2} + B_{1}^{2} + C_{1}^{2})√(A_{2}^{2} + B_{2}^{2} + C_{2}^{2})], where A_{1}i + B_{1}j + C_{1}k , and A_{2}i + B_{2}j + C_{2}k are normal vectors to the two planes, and θ is the angle between the two planes..
Example: Find the angle between two planes with equations 2x + y  2z = 5 and 3x  6y  2z = 7.
Solution: Since the equations of the two planes are given in the cartesian form, we will determine the angle between two planes in cartesian form using the formula cos θ = (A_{1}A_{2} + B_{1}B_{2} + C_{1}C_{2})/[√(A_{1}^{2} + B_{1}^{2} + C_{1}^{2})√(A_{2}^{2} + B_{2}^{2} + C_{2}^{2})]. The equations of the planes are 2x + y  2z = 5 and 3x  6y  2z = 7. Here, A_{1} = 2, B_{1} = 1, C_{1} = 2, A_{2} = 3, B_{2} = 6, C_{2} = 2. Substituting these values into the formula, we have
cos θ = (2×3 + 1×(6) + (2)×(2))/[√(2^{2} + 1^{2} + (2)^{2})√(3^{2} + (6)^{2} + (2)^{2})]
= (6 + (6) + 4)/[√(4 + 1 + 4)√(9 + 36 + 4)]
= 4/(√9 √49)
= 4/(3×7)
= 4/21
⇒ θ = cos^{1}(4/21) [Taking inverse cosine on both sides]
Hence, the angle between the two planes 2x + y  2z = 5 and 3x  6y  2z = 7 is cos^{1}(4/21).
Important Notes on Angle Between Two Planes
 The angle between two planes is equal to the angle between the normal vectors to the two planes and is called the dihedral angle.
 For planes, r.n_{1} = d_{1} and r.n_{2} = d_{2}, the angle between them is given by, cos θ = (n_{1} . n_{2})/(n_{1}.n_{2})
 For planes, A_{1}x + B_{1}y + C_{1}z + D_{1} = 0 and A_{2}x + B_{2}y + C_{2}z + D_{2} = 0, the angle between two planes in cartesian form is given by, cos θ = (A_{1}A_{2} + B_{1}B_{2} + C_{1}C_{2})/[√(A_{1}^{2} + B_{1}^{2} + C_{1}^{2})√(A_{2}^{2} + B_{2}^{2} + C_{2}^{2})]
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Angle Between Two Planes Examples

Example 1: Find the angle between two planes π_{1}: r.(2i  j + k) = 1 and π_{2}: r.(i + k) = 3
Solution: To find the angle between the two planes r.(2i  j + k) = 1 and r.(i + k) = 3, we will use the formula cos θ = (n_{1} . n_{2})/(n_{1}.n_{2}). We have,
n_{1} = 2i  j + k, n_{2} = i + k
n_{1} = √(2^{2} + (1)^{2} + 1^{2}) = √(4 + 1 + 1) = √6
n_{2} = √(1^{2} + (0)^{2} + 1^{2}) = √(1 + 0 + 1) = √2
Scalar product of the normal vectors is given by, n_{1} . n_{2} = (2i  j + k) . (i + k) = 2 × 1 + (1) × (0) + 1 × 1 = 2 + 0 + 1 = 3
Substituting these values into the formula, we have
cos θ = (3)/(√6 . √2)
= 3/√12
= √3/2
⇒ θ = cos^{1}(√3/2)
= π/6 radians
Answer: Hence, the angle between two planes r.(2i  j + k) = 1 and r.(i + k) = 3 is equal to π/6 radians.

Example 2: Determine the angle between two planes P_{1}: 3x  6y + 2z = 7 and P_{2}: 2x + 2y  2z = 3
Solution: We will determine the angle between two planes in cartesian form using the formula cos θ = (A_{1}A_{2} + B_{1}B_{2} + C_{1}C_{2})/[√(A_{1}^{2} + B_{1}^{2} + C_{1}^{2})√(A_{2}^{2} + B_{2}^{2} + C_{2}^{2})]. The equations of the planes are P_{1}: 3x  6y + 2z = 7 and P_{2}: 2x + 2y  2z = 3. Here, A_{1} = 3, B_{1} = 6, C_{1} = 2, A_{2} = 2, B_{2} = 2, C_{2} = 2. Substituting these values into the formula, we have
cos θ = (3×2 + (6)×2 + 2×(2))/[√(3^{2} + (6)^{2} + 2^{2})√(2^{2} + 2^{2} + (2)^{2})]
= (6 + (12)  4)/[√(9 + 36 + 4)√(4 + 4 + 4)]
= 10/(√49 √12)
= 10/(7×2√3)
= 5/7√3
= 5√3/21
⇒ θ = cos^{1}(5√3/21)
Answer: Hence, the angle between the two planes P_{1}: 3x  6y + 2z = 7 and P_{2}: 2x + 2y  2z = 3 is cos^{1}(5√3/21).
FAQs on Angle Between Two Planes
What Is Meant By Angle Between Two Planes?
The angle between two planes is equal to the angle between the normal vectors to the two planes. We can determine the angle between two planes in vector form and cartesian form and is equal to the angle between the normal to the two planes.
What is the Angle Between Two Planes Called?
The angle between two intersecting planes is called a dihedral angle. In other words, we can say that the angle between the normal vectors of two planes is called the dihedral angle.
How to Find Angle Between Two Planes?
We can find the angle between two planes by determining the angle between the normal vectors to the two planes. We can use the following formulas to find the angle between two planes. The equations of the two planes in vector form are r.n_{1} = d_{1} and r.n_{2} = d_{2} and the equations of the two planes in the cartesian form are A_{1}x + B_{1}y + C_{1}z + D_{1} = 0 and A_{2}x + B_{2}y + C_{2}z + D_{2} = 0. Then, the formulas to find the angle θ between two planes are:
 cos θ = (n_{1} . n_{2})/(n_{1}.n_{2})
 cos θ = (A_{1}A_{2} + B_{1}B_{2} + C_{1}C_{2})/[√(A_{1}^{2} + B_{1}^{2} + C_{1}^{2})√(A_{2}^{2} + B_{2}^{2} + C_{2}^{2})]
What is the Dihedral Angle Between Two Planes?
The dihedral angle between two planes is the angle between the two intersecting planes. In simple words, we can say that the angle between the normal vectors of two planes is called the dihedral angle.
How to Find the Cosine of Angle Between Two Planes?
To find the cosine of angle between two planes A_{1}x + B_{1}y + C_{1}z + D_{1} = 0 and A_{2}x + B_{2}y + C_{2}z + D_{2} = 0 in cartesian form and r.n_{1} = d_{1} and r.n_{2} = d_{2} in vector form, we have two formulas:
 cos θ = (n_{1} . n_{2})/(n_{1}.n_{2})
 cos θ = (A_{1}A_{2} + B_{1}B_{2} + C_{1}C_{2})/[√(A_{1}^{2} + B_{1}^{2} + C_{1}^{2})√(A_{2}^{2} + B_{2}^{2} + C_{2}^{2})]
Using these formulas, we can take inverse cosine on both sides and determine the value of θ, the angle between the two planes.
What is the Formula for Angle Between Two Planes in Vector Form?
The formula to find the angle between two planes in vector form r.n_{1} = d_{1} and r.n_{2} = d_{2} is cos θ = (n_{1} . n_{2})/(n_{1}.n_{2}), where θ is the acute angle between the two planes.
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