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Angle between two planes

Angle between two planes - Maths

Various position vectors are placed in the three-dimensional geometry that denotes the location or position of a point concerning the origin. A 3D space has an infinite number of such position vectors. Thus, in a 3D space, we will get a plane with a specific number of position vectors.

Properties of a plane

  • A line can lie within the plane or intersect at a single point or parallel to the plane.
  • Any two distinct planes are either parallel or intersect at a line.
  • Two planes are also parallel to each other if they are perpendicular to the same line.
  • Two lines are parallel to each other if they are perpendicular to the same plane.

How can one calculate the angle between two planes?

The angle between two planes can be calculated with the help of a normal. Thus, we can say that the angle between two planes and the normal is between the planes.

Let n1 and n2 be the two normal to the planes aligned to each other at an angle θ. The equation of two planes can be given by:

r n₂ = d₂

r n₂ = d₂

The angle between two planes is given by-

image

What is a dihedral angle?

A dihedral angle refers to the angle that is between two intersecting planes. It is most commonly used in chemistry. It refers to the angle between planes through two sets of three atoms, which have two atoms in common. In solid geometry, we define it as the union of a line and two half-planes that have this line as a standard edge.

Angle between two planes in a Cartesian plane

Let us consider A₁ x + B₁ y + C₁z + D₁ = 0 and A₂x + B₂y + C₂z + D₂ = 0 be the equation of two planes at an angle θ where A₁, B₁, C₁ and A₂, B₂, C₂ are the direction ratios of the normal to the planes. From the cosine of the angle, we can find the angle between the two planes, and is given by:

image2

Example

Find the angle between the planes whose vector equations are given by r. (2i + 2j – 3k) = 5 and r. (3i – 3j + 5k) = 3.

Solution

Comparing the equation given in the question with the general equation of a plane in vector form, we get,

n₁ = 2i + 2j – 3k and n2 = 3i – 3j + 5k

| n₁ | = (2² + 2² + (-3)²)½ = 17½ and | n₂| = (3² + (-3)² + 5²)½ = 43½

Thus, Cos = (2i + 2j – 3k). (3i – 3j + 5k) / 17½. 43½

Cos = | 2×3 + 2 x (-3) + (-3) x 5 | / 17½. 43½

Cos = | 6 -6 – 15 | / 17½. 43½

Cos = | -15 | / 731½

Cos = 15 / 731½

Therefore, we get the angle as Cos-1 (15 / 731½).

Example

Calculate the angle between two planes given 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

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