Angle between two planes: A plane in geometry is a flat surface that extends in two dimensions indefinitely but has no thickness. The angle formed by the normals of two planes determines the angle formed between them. It can be calculated using the plane’s vector form and cartesian form equations.

The dot product \((\cdot)\) or scalar product of the normal vectors to the two planes can be used to get the angle between them in vector form. The angle between any two planes can be calculated in vector and cartesian form, and it is equal to the angle between the normals of the two planes. The dihedral angle is the angle formed by two planes intersecting.

What is a Plane?

Planes are a crucial component of \(3-\)dimensional geometry. It is a \(2-\)dimensional figure with no thickness that extends infinitely in three-dimensional space. A plane can be thought of as an extended collection of lines organised in three-dimensional space.

Note that in three-dimensional space, an endless number of planes can exist. Position vectors are used in coordinate geometry to show where a point on the plane is in relation to the origin \(\left( {0,0,0} \right)\)

Angle between Two Planes

Similar to the angle between two straight lines or between a straight line and a plane, we can also determine the angle between two planes. By angle between two planes, we actually mean the angle between their normals.

The angle formed by two planes is equal to the angle formed by their normal vectors. We can utilise the scalar product and magnitudes of the normal vectors in the formula to obtain the angle between the two planes since the angle between the normals to these two planes determines the angle between them. Cartesian and vector equations of the plane can be used to calculate the angle between two planes. The equation of a plane in vector form is \(r \cdot n = d\), and its Cartesian equation is \(Ax + By + Cz + D = 0\).

Formulas to Find the Angle Between Two Planes

The angle between two planes is calculated using two formulas. The formulas are available in both vector and cartesian formats.

Angle Between Two Planes in Vector Form

Consider two planes, \({P_1}\) and \({P_2}\) and let the angle between the planes be \({\rm{\theta }}\). The equations of \({P_1}\) and \({P_2}\) in vector form are given by
\(r.{n_1} = {d_1}\)
\(r.{n_2} = {d_2}\)

Now, the cosine of the angle between the planes \({P_1}\) and \({P_2}\) can be found as
\(\cos \theta  = \frac{{\left| {{n_1} \cdot {n_2}} \right|}}{{\left( {\left| {{n_1}} \right| \cdot \left| {{n_2}} \right|} \right)}}\)

where \({n_1}\), and \({n_2}\)are the normal vectors to the two planes \({P_1}\,\& \,{P_2}\)and \({\rm{\theta }}\) is the angle between the two planes.

Angle Between Two Planes in Cartesian Form

Consider two planes, \({P_1}\) and \({P_2}\) and let the angle between the planes be \({\rm{\theta }}\). The equations of \({P_1}\) and \({P_2}\) in the Cartesian form are given by

\({A_1}x + {B_1}y + {C_1}z + {D_1} = 0\)
\({A_2}x + {B_2}y + {C_2}z + {D_2} = 0\)

Now, the cosine of the angle between the planes \({P_1}\) and \({P_2}\) can be found as

\(\cos \theta  = \frac{{\left| {{A_1}{A_2} + {B_1}{B_2} + {C_1}{C_2}} \right|}}{{\left[ {\sqrt {\left( {A_1^2 + B_1^2 + C_1^2} \right)} \sqrt {\left( {A_2^2 + B_2^2 + C_2^2} \right)} } \right]}}\)

where \({A_1}i + {B_1}j + {C_1}k\), and \({A_2}i + {B_2}j + {C_2}k\) are the normal vectors, and \({\rm{\theta }}\) is the angle between the two planes.

We can determine the value of \(\cos \,{\rm{\theta }}\) using the above formulas and then take the cos inverse \(\left( {{{\cos }^{ – 1}}\,{\rm{\theta }}} \right)\) on both sides to get the value of \({\rm{\theta }}\) and hence the angle between two planes.

Angle Bisectors of a Plane

Consider two planes, \({P_1}\) and \({P_2}\) and let the angle between the planes be \({\rm{\theta }}\). The equations of \({P_1}\) and \({P_1}\) in the Cartesian form are given by

\({A_1}x + {B_1}y + {C_1}z + {D_1} = 0\)
\({A_2}x + {B_2}y + {C_2}z + {D_2} = 0\)

The equation of the planes bisecting the angles between these two planes is given by

\(\frac{{{A_1}x + {B_1}y + {C_1}z + {D_1}}}{{\sqrt {\left( {A_1^2 + B_1^2 + C_1^2} \right)} }} =  \pm \frac{{{A_2}x + {B_2}y + {C_2}z + {D_2}}}{{\sqrt {\left( {A_2^2 + B_2^2 + C_2^2} \right)} }}\)

The equation of the plane bisecting the angle between planes
\(r.{n_1} = {d_1}\)
and
\(r.{n_2} = {d_2}\)
in vector form, it is given as
\(\left| {\frac{{r \cdot {n_1} – {d_1}}}{{{n_1}}}} \right| = \left| {\frac{{r \cdot {n_2} – {d_2}}}{{{n_2}}}} \right|\)

Solved Examples – Angle between Two Planes

Q.1. Find the angle between the two planes having equations \(2x + y – 2z = 5\) and \(3x – 6y – 2z = 7\).
Solution: We can see that the equations of the two planes are given in the cartesian form. So, we can determine the angle between the two planes in cartesian form using the formula
\(\cos \theta  = \frac{{\left| {{A_1}{A_2} + {B_1}{B_2} + {C_1}{C_2}} \right|}}{{\left[ {\sqrt {\left( {A_1^2 + B_1^2 + C_1^2} \right)} \sqrt {\left( {A_2^2 + B_2^2 + C_2^2} \right)} } \right]}}\)
The given equations of the planes are \(2x + y – 2z = 5\) and \(3x – 6y – 2z = 7\).
Comparing the given equation with the standard equation of a plane, we get

Variables	\({A_1}\)	\({B_1}\)	\({C_1}\)	\({A_2}\)	\({B_2}\)	\({C_2}\)
Values	\(2\)	\(1\)	\(-2\)	\(3\)	\(-6\)	\(-2\)

Substituting the values into the formula, we get
\(\cos \theta  = \frac{{2 \times 3 + 1 \times ( – 6) + ( – 2) \times ( – 2)}}{{\left[ {\sqrt {\left( {{2^2} + {1^2} + {{( – 2)}^2}} \right)\sqrt {{3^2} + {{( – 6)}^2} + {{( – 2)}^2}} } } \right]}}\)
\( = \frac{{6 + ( – 6) + 4}}{{[\sqrt {(4 + 1 + 4)\sqrt ( 9 + 36 + 4)]} }}\)
\( = \frac{4}{{(\sqrt 9 \sqrt {49} )}}\)
\( = \frac{4}{{(3 \times 7)}}\)
\(\therefore \cos \theta  = \frac{4}{{21}}\)
Now, the value of \({\rm{\theta }}\) can be obtained by taking the cos inverse.
\( \Rightarrow \theta  = {\cos ^{ – 1}}\frac{4}{{21}}\)
Hence, the angle between the given two planes \(2x + y – 2z = 5\) and \(3x – 6y – 2z = 7\) is \({\cos ^{ – 1}}\frac{4}{{21}}\).

Q.2. Find the angle between the planes represented by the vector equations \(r \cdot (2i + 2j – 3k) = 5\) and \(r\cdot (3i – 3j + 5k) = 3\)
Solution: We can see that the equations of the two planes are given in the vector form. So, we can determine the angle between the two planes in vector form using the formula
\(\cos \theta  = \frac{{\left| {{n_1} \cdot {n_2}} \right|}}{{\left( {\left| {{n_1}} \right| \cdot \left| {{n_2}} \right|} \right)}}\)
Compare the given equations with the general equation of a plane in vector form to get,
\({n_1} = 2i + 2j – 3k\) and \({n_2} = 3i – 3j + 5k\)
Now, we have
\(\left| {{n_1}} \right| = \sqrt {\left( {{2^2} + {2^2} + {{( – 3)}^2}} \right)}  = \sqrt {17} \) and \(\left| {{n_2}} \right| = \sqrt {\left( {{3^2} + {{( – 3)}^2} + {5^2}} \right)}  = \sqrt {43} \)
Thus, \(\cos  = \frac{{|(2i + 2j – 3k) \cdot (3i – 3j + 5k)|}}{{\sqrt {17}  \times \sqrt {43} }}\)
\( \Rightarrow \cos \theta  = \frac{{|2 \times 3 + 2 \times ( – 3) + ( – 3) \times 5|}}{{\sqrt {17}  \times \sqrt {43} }}\)
\( \Rightarrow \cos \theta  = \frac{{6 – 6 – 15}}{{\sqrt {17}  \times \sqrt {43} }}\)
\( \Rightarrow \cos \theta  = \frac{{| – 15|}}{{\sqrt {17}  \times \sqrt {43} }}\)
\( \Rightarrow \cos \theta  = \frac{{15}}{{\sqrt {17}  \times \sqrt {43} }}\)
\(\therefore \theta  = {\cos ^{ – 1}}\left( {\frac{{15}}{{\sqrt {17}  \times \sqrt {43} }}} \right)\)
Hence, the angle between the two given planes is \(\theta  = {\cos ^{ – 1}}\left( {\frac{{15}}{{\sqrt {17}  \times \sqrt {43} }}} \right)\).

Q.3. Find the angle between the two planes whose vector equations are given as \(r \cdot (i + j – 3k) = 4\)  and \(r \cdot (i – j + 5k) = 3\).
Solution: We can see that the equations of the planes are given in the vector form. So, we can determine the angle between the two planes in vector form using the formula
\(\cos \theta  = \mid \frac{{{n_1} \cdot {n_2}}}{{\left( {\left| {{n_1}} \right| \cdot \left| {{n_2}} \right|} \right)}}\)
We have
\({n_1} = i + j – 3k\) & \({n_2} = i – j + 5k\)
Now,
\(\left| {{n_1}} \right| = \sqrt {{1^2} + {1^2} + {{( – 3)}^2}}  = \sqrt {1 + 1 + 9}  = \sqrt {11} \)
\(\left| {{n_2}} \right| = \sqrt {{1^2} + {{( – 1)}^2} + {5^2}}  = \sqrt {1 + 1 + 25}  = \sqrt {27} \)
The scalar product of the normal vectors is given by,
\({n_1} \cdot {n_2} = (i + j – 3k) \cdot (i – j + 5k)\)
\( = 1 \times 1 + 1 \times  – 1 + ( – 3) \times 5\)
\( = 1 – 1 – 15\)
\(=-15\)
Substituting the values into the formula, we have
\(\cos \theta  = \frac{{| – 15|}}{{\sqrt {11}  \times \sqrt {27} }}\)
\(\therefore \theta  = {\cos ^{ – 1}}\left( {\frac{{15}}{{\sqrt {11}  \times \sqrt {27} }}} \right)\)
Hence, the angle between the two given planes is \(\theta  = {\cos ^{ – 1}}\left( {\frac{{15}}{{\sqrt {11}  \times \sqrt {27} }}} \right)\).

Q.4. Determine the angle between the following planes: \({P_1} = 2x – y + z – 1 = 0\) and \({P_2} = x + z + 3 = 0\)
Solution: We can see that the equations of the two planes are given in the cartesian form. So, we can determine the angle between the two planes in cartesian form using the formula
\(\cos \theta  = \frac{{\left| {{A_1}{A_2} + {B_1}{B_2} + {C_1}{C_2}} \right|}}{{\left[ {\sqrt {\left. {\left( {A_1^2 + B_1^2 + C_1^2} \right)\sqrt {\left( {A_2^2 + B_2^2 + C_2^2} \right)} } \right]} } \right.}}\)
The given equations of the planes are \({P_1} = 2x – y + z – 1 = 0\) and \({P_2} = x + z + 3 = 0\)
Comparing the given equation with the standard equation of plane, we get
\({A_1} = 2,{B_1} =  – 1,{C_1} = 1,{A_2} = 1,{B_2} = 0,{C_2} = 1\)
Substituting the values into the formula, we get
\(\cos \theta  = \frac{{2 \times 1 +  – 1 \times 0 + 1 \times 1}}{{\left[ {\sqrt {\left( {{2^2} + {{( – 1)}^2} + {{(1)}^2}} \right)\sqrt {{1^2} + {{(0)}^2} + {{(1)}^2}} } } \right]}}\)
\( = \frac{{2 + 0 + 1}}{{[\sqrt {(4 + 1 + 1)\sqrt {(1 + 0 + 1)]} } }}\)
\( = \frac{3}{{(\sqrt {6\sqrt 2 )} }}\)
\( = \frac{3}{{\sqrt {12} }}\)
\( = \frac{3}{{\sqrt {4 \times 3} }} = \frac{3}{{2\sqrt 3 }}\)
\( = \frac{{\sqrt 3  \times \sqrt 3 }}{{2\sqrt 3 }} = \frac{{\sqrt 3 }}{2}\)
Now, the value of \({\rm{\theta }}\) can be obtained by taking the cos inverse.
\( \Rightarrow \theta  = {\cos^{ – 1}} \frac{\sqrt 3}{2} 30^\circ\)
Hence, the angle between the given two planes \({P_1} = 2x – y + z – 1 = 0\) and \({P_2} = x + z + 3 = 0\) is \({\cos ^{ – 1}} \frac{\sqrt 3}{2} 30^\circ \).

Q.5. Find the angle between the planes \(2x + 4y – 4z – 6 = 0\) and \(4x + 3y + 9 = 0\)
Solution: We can see that the equations of the two planes are given in the vector form. So, we can determine the angle between the two planes in vector form using the formula
\(\cos \theta  = \frac{{\left| {{n_1} \cdot {n_2}} \right|}}{{\left( {\left| {{n_1}} \right| \cdot \left| {{n_2}} \right|} \right)}}\)
Therfore,
\(\cos \theta  = \frac{{|2 \cdot 4 + 4 \cdot 3 + ( – 4) \cdot 0|}}{{\sqrt {{2^2} + {4^2} + {{( – 4)}^2}} \sqrt {{4^2} + {3^2} + {0^2}} }}\)
\( = \frac{{8 + 12}}{{\sqrt {36} \sqrt {25} }}\)
\( = \frac{{20}}{{30}} = \frac{2}{3}\)
Hence, the angle between the given two planes is \({\cos ^{ – 1}}\frac{2}{3}\).

Summary

A plane is a two-dimensional flat surface that extends infinitely in two directions. It is formed by a stack of lines that are parallel to each other. Similar to how we see and calculate the angle between two lines, angles will be formed between the two intersecting planes.  

The angle between two planes is the angle formed by the normals of the two planes. According to the geometrical arrangement, the angle between two planes is the same as between their two normal vectors. These normals are perpendicular to the line of intersection of the two planes. The angle between the two planes can be expressed in vector and cartesian forms.

FAQs on Angle Between Two Planes

The frequently asked questions on the angle between two planes are given below:

Q.1. How do you find the angle between two planes?
Ans: The cosine of the angle between the planes \({P_1}\) and \({P_2}\) can be found as
\(\cos \theta  = \frac{{\left| {{n_1} \cdot {n_2}} \right|}}{{\left( {\left| {{n_1}} \right| \cdot \left| {{n_2}} \right|} \right)}}\)
where \({n_1}\), and \({n_1}\) are the normal vectors to the two planes \({P_1}\,\& {P_2}\) and \({\rm{\theta }}\) is the angle between the two planes.

Q.2. What does the angle between two planes mean?
Ans: Similar to the angle between two lines, there is an angle formed between the two intersecting planes.  The angle between the planes is the angle formed by the normals of the two planes. According to the geometrical arrangement, the angle between two planes is the same as between their two normal vectors.

Q.3. What is a plane angle?
Ans: Plane angle is defined as the acute angle between two lines normal to the planes. The angle between a plane and an intersecting straight line is equal to ninety degrees minus the angle between the intersecting line and the line that goes through the point of intersection and is normal to the plane.

Q.4. What is the difference between plane angle and solid angle?
Ans: The solid angle is a three-dimensional geometric entity in a three-dimensional space, while the plane angle is a two-dimensional geometric object on a plane. There are various types of amounts. Comparing the size of the plane and solid angles is similar to comparing the length and area.

Q.5. What is the equation of angle between planes in cartesian form?
Ans: The equations of the plane in the Cartesian form is given by \({A_1}x + {B_1}y + {C_1}z + {D_1} = 0\). Consider two planes, \({P_1}\) and \({P_2}\) and let the angle between the planes be \({\rm{\theta }}\). The equations of \({P_1}\) and \({P_2}\) in the Cartesian form are given by
\({A_1}x + {B_1}y + {C_1}z + {D_1} = 0\)
\({A_2}x + {B_2}y + {C_2}z + {D_2} = 0\)
Now, the cosine of the angle between the planes \({P_1}\) and \({P_2}\) can be found as
\(\cos \theta  = \frac{{\left| {{A_1}{A_2} + {B_1}{B_2} + {C_1}{C_2}} \right|}}{{\left[ {\sqrt {\left( {A_1^2 + B_1^2 + C_1^2} \right)} \sqrt {\left( {A_2^2 + B_2^2 + C_2^2} \right)} } \right]}}\)
where \({A_1}i + {B_1}j + {C_1}k\) , and \({A_2}i + {B_2}j + {C_2}k\) are the normal vectors, and \({\rm{\theta }}\) is the angle between the two planes.
We can determine the value of \(\cos \theta \) using the above formulas and then take the cos inverse \(\left( {{{\cos }^{ – 1}}\,{\rm{\theta }}} \right)\) on both sides to get the value of \(\theta\) and hence the angle between two planes.

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