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126 LearnersLast updated on October 29, 2025

The angle between two planes is the measure of rotation from one plane to another, defined as the acute angle (90°) between their normal vectors. This article discusses the angle between two planes in detail.
In geometry, the angle between two planes is the dihedral angle, which equals the angle between their normals. You can visualize it by drawing perpendicular lines from the line of intersection in each plane; the angle between those lines is the angle between the two planes. Alternatively, working directly with the normal vectors n₁ and n₂, the acute angle θ between the planes satisfies;
\(\cos \theta = \frac{n_1}{n_2} \)
The angle between the two planes is the angle between their normal vectors.
Below are the equations representing the two planes;
Plane 1: \(a_1x + b_1y + c_1z + d_1 = 0 \)
Plane 2: \(a_2x + b_2y + c_2z + d_2 = 0 \)
Now, we extract the normal vector from both planes.
Normal to plane 1: \(\mathbf{n_1} = (a_1,\, b_1,\, c_1) \)
Normal to plane 2:\(\mathbf{n_2} = (a_2,\, b_2,\, c_2) \)
The angle between the planes is the angle between these two vectors.
Use the formula \(\cos \theta = \frac{\mathbf{n_1} \cdot \mathbf{n_2}}{|\mathbf{n_1}|\,|\mathbf{n_2}|} \)
Dot product: \(\mathbf{n_1} \cdot \mathbf{n_2} = a_1a_2 + b_1b_2 + c_1c_2 \)
Magnitude:
\(|\mathbf{n_1}| = \sqrt{a_1^2 + b_1^2 + c_1^2} \)
\(|\mathbf{n_2}| = \sqrt{a_2^2 + b_2^2 + c_2^2} \)
To calculate the angle between two planes, we use the formula
\(\cos \theta = \frac{n_1 n_2}{|n_1|\,|n_2|} \).
As we have already discovered in the previous section, the dot product is
\(n_1 n_2 = a_1a_2 + b_1b_2 + c_1c_2 \) and the magnitude is,
\(n_1 = \sqrt{a_1^2 + b_1^2 + c_1^2}, \quad n_2 = \sqrt{a_2^2 + b_2^2 + c_2^2} \).
Substituting these in the formula, \(\cos \theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \, \sqrt{a_2^2 + b_2^2 + c_2^2}} \)
This is the calculation required to find the angle between two planes in the Cartesian plane.
The angle between two planes is the dot product of the normal vectors of those planes, and can be found using the formula
\(\cos \theta = \frac{n_1 n_2}{|n_1|\,|n_2|} \)
Where,
\(n_1 = (a_1,\, b_1,\, c_1) \)
\(n_2 = (a_2,\, b_2,\, c_2) \)
Angle Between Two Planes in Vector Form
When two planes are written using vector equations, their general form is \(r_n = d \)
Where,
r is the position vector
n is the normal vector of the plane, and
d is a constant.
If,
Plane 1: \(r_{n1} = d_1 \)
Plane 2:\(r_{n2} = d_2 \)
Then the angle between two planes () = the angle between n1and n2.
To determine the angle between two planes, follow the given steps:
The angle between two planes helps determine how they intersect in 3D space. By using the dot product of their normal vectors, you can easily calculate this angle.
Students can miss out on common details while solving for the angle between two planes, which can lead to calculation errors. Here is a list of frequently occurring mistakes and how to fix them.
The geometric concept of angle between two planes helps understand how different objects fit, move, and interact in a three-dimensional space. Given below are some real-world uses of this concept.
Find the angle between the planes x+y+z=1 and 2x+3x+6z=5.
\(\cos \theta = \frac{(1)(2) + (1)(3) + (1)(6)}{\sqrt{(1^2 + 1^2 + 1^2)(2^2 + 3^2 + 6^2)}} = \frac{11}{\sqrt{3 \times 49}} = \frac{11}{7\sqrt{3}} \\ \ \)
\(\theta = \cos^{-1}\left(\frac{11}{7\sqrt{3}}\right) \)
Normals are (1, 1, 1) and (2, 3, 6). Substitute in the formula to get the cosine of the angle between them.
Find the angle between 3x-y+2z=7 and x+2y+2z=9.
\(\cos \theta = \frac{(3)(1) + (-1)(2) + (2)(2)}{\sqrt{(3^2 + (-1)^2 + 2^2)(1^2 + 2^2 + 2^2)}} = \frac{5}{\sqrt{14 \times 9}} = \frac{5}{\sqrt{126}} \)
\(\theta = \cos^{-1}\left(\frac{5}{\sqrt{126}}\right) \)
Use the dot product of the normal vectors (3, -1, 2) and (1, 2, 2) to find the cosine of the angle.
Find the angle between x-y+z=0 and y-z=0.
\(\cos \theta = \frac{(1)(0) + (-1)(1) + (1)(-1)}{\sqrt{(1^2 + (-1)^2 + 1^2)(0^2 + 1^2 + (-1)^2)}} = \frac{-2}{\sqrt{3 \times 2}} = \frac{-2}{\sqrt{6}} \)
\(\theta = \cos^{-1}\left(\frac{-2}{\sqrt{6}}\right) \)
The normal vectors are (1, -1, 1) and (0, 1, -1). Substituting gives the angle between them.
Find the angle between 2x+y-2z=3 and x-3y+6z=4
\(\cos \theta = \frac{(2)(1) + (1)(-3) + (-2)(6)}{\sqrt{(2^2 + 1^2 + (-2)^2)(1^2 + (-3)^2 + 6^2)}} = \frac{-16}{\sqrt{9 \times 46}} = \frac{-16}{\sqrt{414}} \)
\(\theta = \cos^{-1}\left(\frac{-16}{\sqrt{414}}\right) \)
Find the dot product and magnitudes of (2, 1, -2) and (1, -3, 6) to determine the angle.
Find the angle between x+2y+3z=0 and 2x-y+2z=5.
\(\cos \theta = \frac{(1)(2) + (2)(-1) + (3)(2)}{\sqrt{(1^2 + 2^2 + 3^2)(2^2 + (-1)^2 + 2^2)}} = \frac{6}{\sqrt{14 \times 9}} = \frac{6}{\sqrt{126}} \)
\(\theta = \cos^{-1}\left(\frac{6}{\sqrt{126}}\right) \)
The normal vectors are (1, 2, 3) and (2, -1, 2). The formula gives the cosine of the angle between these planes.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.






