Angle between two Planes

In two dimensional space, any two lines can only be either of the following : Intersecting or Parallel.

Similarly, in three dimensional space any given pair of planes can only be either intersecting or parallel.

The region of Intersection of any two intersecting planes is always a line. Consider two intersecting planes P1 & P2. Let letter L represent the line of intersection of the planes, as shown in the picture below.

Imagine a line L1 on plane P1 and a line L2 on plane P2 such that both these lines are perpendicular to the line of intersection L, as depicted below.

The angle between the planes is same as the angle between the lines (L1 & L2) perpendicular to the line of intersection, as seen in the picture above.

In the above picture,

θ

is used to denote angle between lines L1 and L2(and planes). If

θ

is the angle between intersecting lines, then 180° -

θ

is also the angle between the lines, and hence also the angle between the planes. (I have not shown this other angle in the picture).

One of the angles will be acute and the other one will be obtuse, any of them qualifies to be called the angle between the planes.

If an angle between L1 & L2 is 0° or 180°, the planes are parallel.

If an angle between L1 & L2 is 90°, the planes are perpendicular.

In the above picture, the angle between the planes is represented by the symbol

θ

. Let L3 and L4 be lines perpendicular to P1 and P2, and also perpendicular to the lines L1 & L2 , as shown in the picture above.

With a simple bit of geometry, one can easily prove that the angle between the planes is same as the angle between the lines L3 and L4 - lines perpendicular to the planes.

Lines L1, L2, L3 and L4 form a quadrilateral(four-sided figure). And sum of four interior angles of a quadrilateral is 360 degrees. Hence,

90 + 90 + $θ$ + angle between L3 & L4 = 360

angle between L3 & L4 = 180 - $θ$

So the angle between the lines perpendicular to the planes is 180 - $θ$ and $θ$ (external angle of quadrilateral), same as the angle between the planes as shown in the figure above.

So to conclude, the angle between two planes is same as the angle between the lines perpendicular to the two planes which is same as the angle between the lines perpendicular to the line of intersection of the planes.

Let $\vec{d_{1}}$ be some non-zero direction vector of line L3, and $\vec{d_{2}}$ be some non-zero direction vector of L4, as shown below. A direction vector of a line is a vector arrow parallel to that line.

As L3 & L4 are perpendicular to the planes, direction vectors of these lines are also perpendicular to the respective planes. Vectors perpendicular to a given plane are called normal vectors to the plane.

Since the vectors are parallel to L3 & L4, the angle between them will be the same as the angle between L3 & L4. So the angle between them tail-to-tail can be either $θ$ or 180° - $θ$ depending on which side of the plane they are pointing to or away from.

So that means, the angle between the planes is same as the angle between vectors that are normal to the plane.

Note that all we need is normal vectors to find the angle between planes. We don’t necessarily have to refer to them as ‘direction vectors of lines perpendicular to the planes’! It probably would not have made sense if I had told you right at the beginning that the angle between planes is the angle between vectors perpendicular to them. So I needed to elaborate more so that it makes sense as to why it is so.

Example :

P1 : 2x + y - 2z = 5

P2: 3x - 6y - 2z = 7

We want to find the angle between the given planes. The above two are Cartesian equations of planes P1 & P2. Normal vector to the plane is actually “contained” inside its equation. Normal vector to P1 from its equation is $2 \hat{i} + \hat{j} - 2 \hat{k}$ .

Similarly, normal vector to P2 from its equation is $3 \hat{i} - 6 \hat{j} - 2 \hat{k}$ .

Read : How a normal vector helps find the equation of a plane.

Let,

$\vec{d_{1}} = 2 \hat{i} + \hat{j} - 2 \hat{k}$

$\vec{d_{2}} = 3 \hat{i} - 6 \hat{j} - 2 \hat{k}$

So now we need to find the angle between these normal vectors to the planes. The angle between vectors given in component form(in terms of $\hat{i}, \hat{j}, \hat{k}$ ), can be found using either the dot product method or using the cosine rule of triangles method. You can check out this post on how we find the angle between vectors using the dot product method.

The formula for finding the angle between $\vec{d_{1}}$ & $\vec{d_{2}}$ in the above example using the dot product method is,

cos $θ = \frac{\vec{d_{1}} \cdot \vec{d_{2}}}{| \vec{d_{1}} | | \vec{d_{2}} |}$ ….(*)

where,

$θ$ is the angle between vectors,

$\vec{d_{1}} \cdot \vec{d_{2}}$ is the dot product of vectors

$| \vec{d_{1}} |$ & $| \vec{d_{2}} |$ are the magnitudes of vectors

To find the dot product of vectors in the numerator, we first multiply the corresponding x, y and z components, and then add the three results.

$\vec{d_{1}} \cdot \vec{d_{2}}$ = (2)(3) + (1)(-6) + (-2)(-2)

= 6 - 6 + 4

= 4

The magnitude of a vector can be found by squaring the x, y and z components, adding the three squares together, and taking the square root of that.

$\vec{d_{1}} = \sqrt{(2)^{2} + (1)^{2} + (- 2)^{2}}$

$= \sqrt{4 + 1 + 4}$

$= \sqrt{9} = 3$

$\vec{d_{2}} = \sqrt{(3)^{2} + (- 6)^{2} + (- 2)^{2}}$

$= \sqrt{9 + 36 + 4}$

$= \sqrt{49} = 7$

Putting these results back in (*)

cos $θ = \frac{4}{7 \times 3}$

cos $θ = \frac{4}{21}$

The number on the right hand side is cosine of the angle between the vectors. Taking arccosine or inverse cosine of this number will give us the required angle in radians.

$θ = c o s^{-} 1 (\frac{4}{21})$