Angle Between Two Planes

The Angle Between Two Planes is defined as the acute angle created where the planes intersect. It is the angle between the normal vectors of the planes. Mathematically, if and are the normal vectors of the two planes, the angle θ between the planes can be calculated using the dot product:

$\cos \theta =\frac{{\overrightarrow{\mathbf{n}}}_1\cdot {\overrightarrow{\mathbf{n}}}_2}{|{\overrightarrow{\mathbf{n}}}_1||{\overrightarrow{\mathbf{n}}}_2|}$

where ${\overrightarrow{\mathbf{n}}}_1\cdot {\overrightarrow{\mathbf{n}}}_2$ is the dot product of the normal vectors, and $\left|{\overrightarrow{\mathbf{n}}}_1\right|$ and $\left|{\overrightarrow{\mathbf{n}}}_2\right|$ are the magnitudes of the normal vectors. This angle is typically measured in degrees or radians.

1.0Angle Between Two Planes in Vector Form

Consider two planes intersecting each other in a straight line.

If ${\overrightarrow{\mathrm{n}}}_1$ is normal to one plane and ${\overrightarrow{\mathrm{n}}}_2$ is normal to the other plane then the planes are described by the equation $\overrightarrow{\mathrm{r}}\cdot {\overrightarrow{\mathrm{n}}}_1={\mathrm{d}}_1$ and $\overrightarrow{\mathrm{r}}\cdot {\overrightarrow{\mathrm{n}}}_2={\mathrm{d}}_2$.

Let us assume the acute angle formed between two planes is θ . Obtuse angle formed Between two planes is 180 – θ.

So, the angle between the normal is θ. Let us now calculate angle between the 2 normal which can be calculated by the fact that the dot product of the vectors ${\overrightarrow{\mathrm{n}}}_1$ and ${\overrightarrow{\mathrm{n}}}_2$ is equal to the product of magnitude and cosine of angle θ.

${\overrightarrow{\mathrm{n}}}_1\cdot {\overrightarrow{\mathrm{n}}}_2=\left|{\overrightarrow{\mathrm{n}}}_1\right|\left|{\overrightarrow{\mathrm{n}}}_2\right|\cos \theta$

$\frac{{\overrightarrow{\mathrm{n}}}_1\cdot {\overrightarrow{\mathrm{n}}}_2}{|{\overrightarrow{\mathrm{n}}}_1||{\overrightarrow{\mathrm{n}}}_2|}=\cos \theta$

$\theta ={\cos }^{-1}\left|\frac{{\overrightarrow{\mathrm{n}}}_1\cdot {\overrightarrow{\mathrm{n}}}_2}{|{\overrightarrow{\mathrm{n}}}_1||{\overrightarrow{\mathrm{n}}}_2|}\right|$

Note: ${\overrightarrow{\mathrm{n}}}_1\times {\overrightarrow{\mathrm{n}}}_2$ is parallel to the intersection of 2 planes.

Important Points:

• The line of intersection of planes is always perpendicular to both the normal.

• If the normal ${\overrightarrow{\mathrm{n}}}_1$ and ${\overrightarrow{\mathrm{n}}}_2$ are perpendicular to each other that means they are at 90° with each other, then dot product is equal to 0. 

${\overrightarrow{\mathrm{n}}}_1\cdot {\overrightarrow{\mathrm{n}}}_2=\left|{\overrightarrow{\mathrm{n}}}_1\right|cdot\left|{\overrightarrow{\mathrm{n}}}_2\right|\cos 90^{\circ }$

${\overrightarrow{\mathrm{n}}}_1\cdot {\overrightarrow{\mathrm{n}}}_2=0\left\{\because \cos 90^{\circ }=0\right\}$

• If the dot product of ${\overrightarrow{\mathrm{n}}}_1$ and ${\overrightarrow{\mathrm{n}}}_2$ are zero, then the planes are perpendicular to each other, and the angle defined by the planes can be defined as the angle between the normal.
• If the angle between the normal ${\overrightarrow{\mathrm{n}}}_1$ and ${\overrightarrow{\mathrm{n}}}_2$ is 0° then the normal are parallel to each other. 
• Two planes are parallel to each other if their normal are parallel. 

Summary:

• Angle between the two planes is defined as the angle between their normal.

• If θ is the angle between the two planes represented by the equations $\overrightarrow{\mathrm{r}}\cdot {\overrightarrow{\mathrm{n}}}_1={\mathrm{d}}_1$ and $\overrightarrow{\mathrm{r}}\cdot {\overrightarrow{\mathrm{n}}}_2={\mathrm{d}}_2$ ,  then $\theta ={\cos }^{-1}\left|\frac{{\overrightarrow{\mathrm{n}}}_1\cdot {\overrightarrow{\mathrm{n}}}_2}{|{\overrightarrow{\mathrm{n}}}_1||{\overrightarrow{\mathrm{n}}}_2|}\right|.$

• Two planes are perpendicular to each other if dot product of their normal is equal to zero.

• Two planes are parallel if their normal are parallel to each other.

2.0Angle Between Two Planes in Coordinate Form

Angle between the planes ${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1\mathrm{z}+{\mathrm{d}}_1=0$ and ${\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2\mathrm{z}+{\mathrm{d}}_2=0$ is

$\cos \theta =\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2}{\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2+{\mathrm{c}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2+{\mathrm{c}}_2^2}}$

If direction cosine is given, then

$\cos \theta =\frac{{\ell }_1{\ell }_2+{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{n}}_1{\mathrm{n}}_2}{\sqrt{{\ell }_1^2+{\mathrm{m}}_1^2+{\mathrm{n}}_1^2}\sqrt{{\ell }_2^2+{\mathrm{m}}_2^2+{\mathrm{n}}_2^2}}$

$\cos \theta ={\ell }_1{\ell }_2+{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{n}}_1{\mathrm{n}}_1$

$\left[\begin{array}{l}{\ell }_1^2+{\mathrm{m}}_1^2+{\mathrm{n}}_1^2=1\\ {\ell }_2^2+{\mathrm{m}}_2^2+{\mathrm{m}}_2^2=1\end{array}\right]$

If θ = 90°, then cos 90° = 0 (two planes are perpendicular)

∴ a₁a₂ + b₁b₂ + c₁c₂ = 0 and ${\ell }_1{\ell }_2+{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{n}}_1{\mathrm{n}}_2=0$

If θ = 0°, $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$ (two planes are parallel)

3.0Solved Examples on Angle Between Two Planes

Example 1: The angle between the plane 2x – y + z = 6 and x + y + 2z = 3 is

Solution:

a₁ = 2, b₁ = –1, c₁ = 1

a₂ = 1, b₂ = 1, c₂ = 2

Comparing both equation by ${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1\mathrm{z}+{\mathrm{d}}_1=0$ and ${\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2\mathrm{z}+{\mathrm{d}}_2=0$

Using formula for cartesian forms

$\cos \theta =\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2}{\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2+{\mathrm{c}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2+{\mathrm{c}}_2^2}}$

$\Rightarrow \cos \theta =\frac{2\cdot 1+(-1)\cdot 1+1\cdot 2}{\sqrt{2^2+(-1{)}^2+1^2}\sqrt{1^2+1^2+2^2}}$

$\Rightarrow \cos \theta =\frac{2-1+2}{\sqrt{4+1+1}\sqrt{1+1+4}}$

$\Rightarrow \cos \theta =\frac{3}{\sqrt{6}\sqrt{6}}$

$\Rightarrow \cos \theta =\frac{3}{6}$

$\Rightarrow \cos \theta =\frac{1}{2}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{2}\right)$

$\Rightarrow \theta =\frac{\pi }{3}$

Example 2: Find the angle between the planes 2x – 3y + 4z = 5 and 3x + 5y – 2z = 7

Solution:

Comparing both equations by ${\mathrm{a}}_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1\mathrm{z}-{\mathrm{d}}_1=0$ and ${\mathrm{a}}_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2\mathrm{z}-{\mathrm{d}}_2=0$.

So, ${\mathrm{a}}_1$=2, ${\mathrm{b}}_1$=-3, ${\mathrm{c}}_1$=4

and a2 = 3, b2 = 5, c2 = –2

Using formula for cartesian form

$\cos \theta =\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

$\Rightarrow \cos \theta =\frac{2\cdot 3+(-3)\cdot 5+4\cdot (-2)}{\sqrt{2^2+(-3{)}^2+4^2}\sqrt{3^2+5^2+(-2{)}^2}}$

$\Rightarrow \cos \theta =\frac{6-15-8}{\sqrt{4+9+16}\sqrt{9+25+4}}$

$\Rightarrow \cos \theta =\frac{-17}{\sqrt{29}\sqrt{38}}$

$\Rightarrow \cos \theta =\frac{-17}{\sqrt{1102}}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{-17}{\sqrt{1102}}\right)$

Example 3: Find the angle between the planes 3x + y – 2z = 7 and 2x – 4y + z = 3.

Solution:

Comparing both equations by $a_1\mathrm{x}+{\mathrm{b}}_1\mathrm{y}+{\mathrm{c}}_1\mathrm{z}-{\mathrm{d}}_1=0$ and

$a_2\mathrm{x}+{\mathrm{b}}_2\mathrm{y}+{\mathrm{c}}_2\mathrm{z}-{\mathrm{d}}_2=0$.

So, a₁ = 3, b₁ = 1, c₁ = –2 

and a₂ = 2, b₂ = –4, c₂ = 1

Using formula for cartesian form

$\cos \theta =\frac{{\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{b}}_1{\mathrm{b}}_2+{\mathrm{c}}_1{\mathrm{c}}_2}{\sqrt{{\mathrm{a}}_1^2+{\mathrm{b}}_1^2+{\mathrm{c}}_1^2}\sqrt{{\mathrm{a}}_2^2+{\mathrm{b}}_2^2+{\mathrm{c}}_2^2}}$

$\Rightarrow \cos \theta =\frac{3\cdot 2+1(-4)+(-2)1}{\sqrt{3^2+1^2+(-2{)}^2}\sqrt{2^2+(-4{)}^2+1^2}}$

$\Rightarrow \cos \theta =\frac{6-4-2}{\sqrt{9+1+4}\sqrt{4+16+1}}$

$\Rightarrow \cos \theta =\frac{0}{\sqrt{14}\sqrt{21}}$

$\Rightarrow \cos \theta =0$

$\Rightarrow \theta ={\cos }^{-1}(0)$

$\Rightarrow \theta =90^{\circ }$

So, the angle between the planes is $90^{\circ }$.

Example 4: Find the angle between the planes 2x + 3y – z = 5 and x – 4y + 2z = –6.

Solution:

For the plane 2x + 3y – z = 5, the normal vector ${\overrightarrow{\mathrm{n}}}_1=⟨2,3,-1⟩$

For the plane x – 4y + 2z = –6, the normal vector ${\overrightarrow{\mathrm{n}}}_2=⟨1,-4,2⟩$

Using formula

$\cos \theta =\frac{{\overrightarrow{\mathrm{n}}}_1\cdot {\overrightarrow{\mathrm{n}}}_2}{|{\overrightarrow{\mathrm{n}}}_1||{\overrightarrow{\mathrm{n}}}_2|}$

$\Rightarrow \cos \theta =\frac{2\cdot 1+3(-4)+(-1)\cdot 2}{\sqrt{2^2+3^2+(-1{)}^2}\sqrt{1^2+(-4{)}^2+2^2}}$

$\Rightarrow \cos \theta =\frac{2-12-2}{\sqrt{4+9+1}\sqrt{1+16+4}}$

$\Rightarrow \cos \theta =\frac{-12}{\sqrt{14}\sqrt{21}}$

$\Rightarrow \cos \theta =\frac{-12}{\sqrt{29}}$

$\Rightarrow \cos \theta =\frac{-12}{7\sqrt{6}}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{-12}{7\sqrt{6}}\right)$

Example 5: Find the angle between the planes x – 2y + 3z = 4 and 4x + y – z = 2.

Solution:

For the plane x – 2y + 3z = 4, the normal vector ${\overrightarrow{\mathrm{n}}}_1=⟨1,-2,3⟩$

For the plane 4x + y – z = 2, the normal vector ${\overrightarrow{\mathrm{n}}}_2=⟨4,1,-1⟩$

Dot Product of both normal are:

${\overrightarrow{\mathrm{n}}}_1\cdot {\overrightarrow{\mathrm{n}}}_2$= (1)(4) + (–2)(1) + (3)(–1)

= 4 – 2 – 3 = 1

Magnitude of normal are:

$\left|{\vec{n}}_1\right|=\sqrt{1^2+(-2{)}^2+3^2}=\sqrt{1+4+9}=\sqrt{14}$

$\left|{\overrightarrow{\mathrm{n}}}_2\right|=\sqrt{4^2+1^2+(-1{)}^2}=\sqrt{16+1+1}=\sqrt{18}=3\sqrt{2}$

Using formula

$\cos \theta =\frac{{\overrightarrow{\mathrm{n}}}_1\cdot {\overrightarrow{\mathrm{n}}}_2}{|{\overrightarrow{\mathrm{n}}}_1|\cdot |{\overrightarrow{\mathrm{n}}}_2|}$

$\Rightarrow \cos \theta =\frac{-1}{\sqrt{14}\cdot 3\sqrt{2}}$

$\Rightarrow \cos \theta =\frac{-1}{3\sqrt{28}}$

$\Rightarrow \cos \theta =\frac{-1}{3\cdot 2\sqrt{7}}$

$\Rightarrow \cos \theta =\frac{-1}{6\sqrt{7}}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{-1}{6\sqrt{7}}\right)$

4.0Practice Questions on Angle Between Two Planes

1. Find the angle between the planes 3x + 2y – z = 0 and x – y + z = 1.

2. Find the angle between the planes 2x + 3y – z = 5 and x – 4y + 2z = –6.

3. Find the angle between the planes 4x + y + z = 1 and x – y + z = 2.

4. Find the angle between the planes 3x + 2y – z = 6 and x – 4y + 2z = 7.

5. Find the angle between the planes 5x – 3y + 2z = 8 and 3x + y – z = 4.

5.0Frequently Asked Questions on Angle Between Two Planes

1. How do you find the angle between two planes?

Ans: To find the angle between two planes, you can use the dot product of their normal vectors. If ${\mathbf{n}}_1$ and ${\mathbf{n}}_2$ are the normal vectors of the planes, the angle q can be calculated using:$\cos \theta =\frac{{\mathbf{n}}_1\cdot {\mathbf{n}}_2}{|{\mathbf{n}}_1||{\mathbf{n}}_2|}$. Then, $\theta ={\cos }^{-1}(\cos \theta ).$

2. What are normal vectors of a plane?

Ans: Normal vectors are vectors that are perpendicular to the plane. For a plane given by ax + by + cz = d, the normal vector is $\mathbf{n}=⟨a,b,c⟩$.

3. How do you find the direction cosines of a normal vector?

Ans: Direction cosines are the cosines of the angles that a vector makes with the coordinate axes. For a vector $\mathbf{n}=⟨a,b,c⟩$, the direction cosines are:

$\mathrm{l}=\frac{\mathrm{a}}{|\mathrm{n}|},\mathrm{m}=\frac{\mathrm{b}}{|\mathrm{n}|},\mathrm{n}=\frac{\mathrm{c}}{|\mathrm{n}|}$, where, |n|=$\sqrt{a^2+b^2+c^2}$ .