Angle Between a Line and a Plane: Definition, Formula

Angle Between a Line and a Plane: The geometry which defines a position of a point in space by three numbers x, y and z is called three-dimensional coordinate geometry or analytical geometry. The angle between a line and a plane is the complement of the angle between the line and the normal plane.

Let us first define some key definitions like line, plane, normal to a plane and equation of a plane in different forms, which help us understand the concept angle between a line and a plane.

Angle Between a Line and a Plane Definition

A surface is a plane if every point on the line joining any two arbitrary points on the surface lies on the surface. This is Euclid’s definition of a plane. The position of a plane in space can be determined in four ways.

• A line, and a point that is not lying on the line
• Three points are not lying on one line
• Two intersecting lines
• Two parallel lines

Each of these methods can be used to find the position and hence, the equation of a plane in space.

What is Normal to a Plane?

A plane can be thought of as made of a point \(P\left( {{x_1},{y_1},{z_1}} \right)\) and all other points \(Q\) such that the segment \(PQ\) is perpendicular to a given direction \(a, b, c\) called the normal direction to the plane.

General Form of a Plane

A first-degree equation in \(x, y, z\) represents a plane. The general equation of a plane is \(ax + by + cz + d = 0\)

where \(a, b\) and \(c\) are constants.

The general equation of a plane passing through a given point \(\left( {{x_1},{y_1},{z_1}} \right)\) is \(a\left( {x – {x_1}} \right) + b\left( {y – {y_1}} \right) + c\left( {z – {z_1}} \right) = 0\)

where \(a, b\) and \(c\) are constants.

Equation of Plane in Normal Form

Let us know about the equation of plane in normal form in the below table:

Type	Description	Equation
Vector Form		\(\vec r \cdot \hat n = d\)
Cartesian Form	Let \(P(x, y, z)\) be any point on the plane	\(lx + my + nz = d\) where \(l, m,\) and \(n\) be the direction cosines of \(\hat n\).

Note

1. If \(\overrightarrow r .\left( {a\hat i + b\hat j + c\hat k} \right) = d\) is the vector equation of a plane, then \(ax + by + cz = d\) is the cartesian equation of the plane, where \(a, b,\) and \(c\) are the direction ratios of the normal to the plane.

If in the equation \(\overrightarrow r .\overrightarrow n = d,\,\overrightarrow n \) is not a unit vector, then we reduce the equation \(\vec r \cdot \vec n = d\) to normal form by dividing both sides by \(|\hat n|\), we get \(\frac{{\vec r \cdot \vec n}}{{|\hat n|}} = \frac{d}{{|\hat n|}}\)

Equation of a Plane Normal to a Given Vector and Passing Through a Given Point

Here is a tabular representation for deeper understanding:

Equation of Plane Passing with Three Given Points

Equation of plane passing with three given points can be described in the following manner:

Angle Between Line and Plane

The angle between a line and a plane is the complement of the angle between the line and the normal to the plane.

Let \(\alpha \) be the plane and \(L\) the line which is not parallel to \(\alpha \). Let the line \(L\) make an angle \(\theta \left( {0 < \theta < \frac{\pi }{2}} \right)\) with the plane \(\alpha \). Then \(\left( {\frac{\pi }{2} – \theta } \right)\) is the acute angle between the normal to the plane and the line \(L\).

Vector Form

Let the given line be \(\vec r = \vec a + \lambda \vec b\) and the given plane be \(\vec r \cdot \vec n = d\).

If \(\phi \) is the angle between the given line (parallel to \(\vec b\)) and the normal to the plane. So, \(\vec n \cdot \vec b = |\vec n||\vec b|\cos \phi = nb\cos \phi \)

If \(\theta \) is the angle between the line and plane, then \(\theta = \left( {\frac{\pi }{2} – \emptyset } \right) = \left( {\frac{\pi }{2} – \left( {\frac{{\vec n \cdot \vec b}}{{nb}}} \right)} \right) = \left( {\frac{{\vec n \cdot \vec b}}{{nb}}} \right)\).

Cartesian Form

Let the given line be \(\frac{{x – {x_1}}}{l} = \frac{{y – {y_1}}}{m} = \frac{{z – {z_1}}}{n}\) and the given plane be \(ax + by + cz + d = 0\).

The direction ratios of the line are \(l, m, n\). The equation of the plane is given to be \(ax + by + cz + d = 0\) so that the direction ratios of the normal to the plane are \(a, b, c\). Suppose that \(\theta \) be the angle between the line and the plane where \(\theta \) is an acute angle. Then \(\left( {\frac{\pi }{2} – \theta } \right)\) is the acute angle between the given line and the normal to the given plane.

Hence, \(\cos \left( {\frac{\pi }{2} – \theta } \right) = \frac{{|al + bm + cn|}}{{\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{a^2} + {b^2} + {c^2}} }}\)

\( \Rightarrow \sin \theta = \frac{{|al + bm + cn|}}{{\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{a^2} + {b^2} + {c^2}} }}\)

\( \Rightarrow \theta = \left[ {\frac{{|al + bm + cn|}}{{\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{a^2} + {b^2} + {c^2}} }}} \right]\) ; where \(\left( {0 < \theta < \frac{\pi }{2}} \right)\).

Which gives the required angle between the line and the plane.

Conditions of Parallelism and Perpendicularity

The following are few conditions of parallelism and Perpendicularity:

Conditions of Parallelism

If a line is parallel to the plane, then normal to the plane is perpendicular to the given line (i.e. \(\theta = 0\))

\( \Rightarrow al + bm + cn = 0\)

Also, since \(\left( {{x_1},{y_1},{z_1}} \right)\) does not lie in the plane \(ax + by + cz + d = 0\).

\(\therefore \,\,a{x_1} + b{y_1} + c{z_1} + d \ne 0\)

Conditions of Perpendicularity

If a line is perpendicular to the plane (i.e. \(\theta = \frac{\pi }{2}\)) then the line is parallel to the normal to the plane. But the direction ratios of the line are \(l, m, n\) and the direction ratios of the normal to the plane are \(a, b, c\). Hence \(\frac{l}{a} = \frac{m}{b} = \frac{n}{c}\) is required conditions for perpendicular.

Point of Intersection

To find the point of intersection of the line \(\frac{{x – {x_1}}}{l} = \frac{{y – {y_1}}}{m} = \frac{{z – {z_1}}}{n}\) and the plane be \(ax + by + cz + d = 0\)

Let \(\frac{{x – {x_1}}}{l} = \frac{{y – {y_1}}}{m} = \frac{{z – {z_1}}}{n} = k\)

\( \Rightarrow x = {x_1} + lk,y = {y_1} + mk,z = {z_1} + nk\,\,\,\,\,…(1)\)

If the point \((x, y, z)\) lies on the plane \(ax + by + cz = d\), then \(a\left( {{x_1} + lk} \right) + b\left( {{y_1} + mk} \right) + c\left( {{z_1} + nk} \right) + d = 0\)

\( \Rightarrow k = \frac{{a{x_1} + b{y_1} + c{z_1} + d}}{{al + bm + cn}}\)

Putting the value of \(k\) in \((1)\) we get the coordinates of the point of intersection. Here we suppose that the line is not parallel to the plane so that \(al + bm + cn \ne 0\).

Solved Examples of Angle Between a Line and a Plane

Here are a few solved examples that can give better insight:

Q.1. Find the angle between the line \(\vec r = (\hat i + 2\hat j – \hat k) + \lambda (\hat i – \hat j + \hat k)\) and the plane \(\vec r \cdot (2\hat i – \hat j + \hat k) = 4\).

Ans: We know that \(\theta \) is the angle between the line be \(\vec r = \vec a + \lambda \vec b\) and plane be \(\vec r \cdot \vec n = d\), then \(\sin \theta = \left| {\frac{{\vec n \cdot \vec b}}{{|n||b|}}} \right|\)

Therefore, if \(\theta \) is the angle between \(\vec r = (\hat i + 2\hat j – \hat k) + \lambda (\hat i – \hat j + \hat k)\) and \(\vec r.(2\hat i – \hat j+ \hat k) = 4\).

Here, \(\vec b = (\hat i – \hat j + \hat k)\), and \(\vec n = (2\hat i – \hat j + \hat k)\)

Then, \(\sin \theta = \frac{{(\hat i – \hat j + \hat k) \cdot (2\hat i – \hat j + \hat k)}}{{|\hat i – \hat j + \hat k||2\hat i – \hat j + \hat k|}} = \frac{{2 + 1 + 1}}{{\sqrt {1 + 1 + 1} \sqrt {4 + 1 + 1} }} = \frac{4}{{\sqrt 3 \sqrt 6 }} = \frac{4}{{3\sqrt 2 }}\)

\( \Rightarrow \theta = \frac{4}{{3\sqrt 2 }}\)

Hence, the angle between \(\vec r = (\hat i + 2\hat j – \hat k) + \lambda (\hat i – \hat j + \hat k)\) and \(\vec r \cdot (2\hat i – \hat j + \hat k) = 4\) is \(\frac{4}{{3\sqrt 2 }}\).

Q.2. Find the angle between the line \(\frac{{x + 1}}{2} = \frac{y}{3} = \frac{{z – 3}}{6}\) and the plane \(3x + y + z = 7\).

Ans: Let \(\theta \) be the angle between the line and the plane. Then \(\left( {\frac{\pi }{2} – \theta } \right)\) will be the angle between the line and the normal to the plane.

\(\cos \left( {\frac{\pi }{2} – \theta } \right) = \sin (\theta ) = \frac{{|al + bm + cn|}}{{\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{a^2} + {b^2} + {c^2}} }}\)

\(\sin (\theta ) = \frac{{2 \times 3 + 3 \times 1 + 6 \times 1}}{{\sqrt {{2^2} + {3^2} + {6^2}} \sqrt {{3^2} + {1^2} + {1^2}} }}\)

\(\sin (\theta ) = \frac{{6 + 3 + 6}}{{\sqrt {49} \sqrt {11} }} = \frac{{15}}{{7\sqrt {11} }}\)

Therefore, \(\theta = \frac{{15}}{{7\sqrt {11} }}\).

Q.3. Find the angle between the line \(\frac{{x + 2}}{3} = \frac{y}{4} = \frac{{z – 2}}{3}\) and the plane \(4x + 2y + 2z = 9\).

Ans: Let \(\theta \) be the angle between the line and the plane. Then \(\left( {\frac{\pi }{2} – \theta } \right)\) will be the angle between the line and the normal to the plane.

\(\cos \left( {\frac{\pi }{2} – \theta } \right) = \sin (\theta ) = \frac{{|al + bm + cn|}}{{\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{a^2} + {b^2} + {c^2}} }}\)

\(\sin (\theta ) = \frac{{3 \times 4 + 4 \times 2 + 3 \times 2}}{{\sqrt {{3^2} + {4^2} + {3^2}} \sqrt {{4^2} + {2^2} + {2^2}} }}\)

\(\sin (\theta ) = \frac{{12 + 8 + 6}}{{\sqrt {34} \sqrt {24} }} = \frac{{26}}{{4\sqrt {51} }} = \frac{{13}}{{2\sqrt {51} }}\)

Therefore, \(\theta = \frac{{13}}{{2\sqrt {51} }}\).

Q.4. Find the angle between the line \(\vec r = (2\hat i + \hat j – 3\hat k) + \lambda (2\hat i – 2\hat j + 3\hat k)\) and the plane \(\vec r \cdot (3\hat i – 2\hat j + 2\hat k) = 5\).

Ans: We know that \(\theta \) is the angle between the line be \(\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b \) and plane be \(\vec r \cdot \vec n = d\), then \(\sin \theta = \left| {\frac{{\vec n \cdot \vec b}}{{|n||b|}}} \right|\)

Therefore, if \(\theta \) is the angle between the line \(\vec r = (2\hat i + \hat j – 3\hat k) + \lambda (2\hat i – 2\hat j + 3\hat k)\) and the plane \(\vec r \cdot (3\hat i – 2\hat j + 2\hat k) = 5\).

Here, \(\vec b = (2\hat i – 2\hat j + 3\hat k)\), and \(\vec n = (3\hat i – 2\hat j + 2\hat k)\)

Then, \(\sin \theta = \frac{{(3\hat i – 2\hat j + 2\hat k) \cdot (2\hat i – 2\hat j + 3\hat k)}}{{|3\hat i – 2\hat j + 2\hat k||2\hat i – 2\hat j + 3\hat k|}} = \frac{{6 + 4 + 6}}{{\sqrt {9 + 4 + 4} \sqrt {4 + 4 + 9} }} = \frac{{16}}{{\sqrt {17} \sqrt {17} }} = \frac{{16}}{{17}}\)

\( \Rightarrow \theta = \frac{{16}}{{17}}\)

Hence, the angle the between \(\vec r = (2\hat i + \hat j – 3\hat k) + \lambda (2\hat i – 2\hat j + 3\hat k)\) and the plane \(\vec r.(3\hat i – 2\hat j + 2\hat k) = 5\) is \(\frac{{16}}{{17}}\).

Q.5. Find the angle between the line \(\vec r = (\hat i + \hat k) + \lambda (\hat j + \hat k)\) and the plane \(\vec r \cdot (2\hat i – \hat j ) = 3\).

Ans: We know that \(\theta \) is the angle between the line be \(\vec r = \vec a + \lambda \vec b\) and plane be \(\vec r \cdot \vec n = d\), then \(\sin \theta = \left| {\frac{{\vec n \cdot \vec b}}{{|n||b|}}} \right|\)

Therefore, if \(\theta \) is the angle between the line \(\vec r = (\hat i + \hat k) + \lambda (\hat j + \hat k)\) and the plane \(\vec r.(2\hat i – \hat j ) = 3\).

Here, \(\vec b = (\hat j + \hat k)\), and \(\vec n = (2\hat i – \hat j )\)

Then, \(\sin \theta = \frac{{(2\hat i – \hat j) \cdot (\hat j + \hat k)}}{{|(2\hat i – \hat j)||(\hat j + \hat k)|}} = \frac{{0 – 1 + 0}}{{\sqrt {4 + 1} \sqrt {1 + 1} }} = \frac{{ – 1}}{{\sqrt 5 \sqrt 2 }} = \frac{{ – 1}}{{\sqrt {10} }}\)

\( \Rightarrow \theta = \left( {\frac{{ – 1}}{{\sqrt {10} }}} \right)\)

Hence, the angle between the line \(\vec r = (\hat i + \hat k) + \lambda (\hat j + \hat k)\) and the plane \(\vec r \cdot (2\hat i – \hat j ) = 3\) is \(\left( {\frac{{ – 1}}{{\sqrt {10} }}} \right)\).

Summary

The angle between a line and a plane is the complement of the angle between the line and the normal plane. If \(\theta \) is the angle between the line \(\vec r = \vec a + \lambda \vec b\) and plane \(\vec r.\vec n = d\), then \(\theta = \left( {\frac{{\vec n \cdot \vec b}}{{nb}}} \right)\). Let \(\theta \) be the angle between the line \(\frac{{x – {x_1}}}{l} = \frac{{y – {y_1}}}{m} = \frac{{z – {z_1}}}{n}\) and the given plane \(ax + by + cz + d = 0\) then \(\theta = \left[ {\frac{{|al + bm + cn|}}{{\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{a^2} + {b^2} + {c^2}} }}} \right]\) ; where \(\left( {0 < \theta < \frac{\pi }{2}} \right)\). If the line is parallel to the normal of the plane, then the line is perpendicular to the plane. If the line is parallel to the given plane, then normal to the plane is perpendicular to the given line i.e. \(al + bm + cn = 0\).

Frequently Asked Questions (FAQs)

Students must be having many questions regarding Angle Between a Line and a Plane. Here are a few commonly asked questions and answers:

Q.1. How to find the acute angle between a line and a plane?
Ans: The angle between a line and a plane is the complement of the angle between the line and the normal to the plane. Let \(\alpha \) be the plane and \(L\) the line which is not parallel to \(\alpha \). Let the line \(L\) make an angle \(\theta \left( {0 < \theta < \frac{\pi }{2}} \right)\) with the plane \(\alpha \). Then \(\left( {\frac{\pi }{2} – \theta } \right)\) is the acute angle between the normal to the plane and the line \(L\).

Q.2. Find the angle between a line and a plane?
Ans: Let the given line be \(\vec r = \vec a + \lambda \vec b\) and the given plane be \(\vec r \cdot \vec n = d\) and \(\theta \) is the angle between the line and plane, then \(\theta = \left( {\frac{\pi }{2} – \left( {\frac{{\vec n \cdot \vec b}}{{nb}}} \right)} \right) = \left( {\frac{{\vec n \cdot \vec b}}{{nb}}} \right)\)

Q.3. How do you find the angle between a line and a plane?
Ans: Let the given line be \(\vec r = \vec a + \lambda \vec b\) and the given plane be \(\vec r \cdot \vec n = d\).
If \(\phi \) is the angle between the given line (parallel to \(\vec b\)) and the normal to the plane. So, \(\vec n \cdot \vec b = |\vec n||\vec b|\cos \phi = nb\cos \phi \)
\( \Rightarrow \cos \phi = \frac{{\vec n \cdot \vec b}}{{nb}} \Rightarrow \left( {\frac{{\vec n \cdot \vec b}}{{nb}}} \right)\)

Q.4. How to find the angle between a line and a plane?
Ans: Let the given line be \(\frac{{x – {x_1}}}{l} = \frac{{y – {y_1}}}{m} = \frac{{z – {z_1}}}{n}\) and the given plane be \(ax + by + cz + d = 0\). If \(\phi \) is the angle between the given line and the normal to the plane then \(\emptyset = \left[ {\frac{{|al + bm + cn|}}{{\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{a^2} + {b^2} + {c^2}} }}} \right]\)

Q.5. What is the formula for the angle between a line and a plane?
Ans: Let the given line be \(\vec r = \vec a + \lambda \vec b\) and the given plane be \(\vec r \cdot \vec n = d\).
If \(\phi \) is the angle between the given line (parallel to \(\vec b\)) and the normal to the plane. So, \(\vec n \cdot \vec b = |\vec n||\vec b|\cos \phi = nb\cos \phi \)
\( \Rightarrow \cos \phi = \frac{{\vec n \cdot \vec b}}{{nb}}\)
\(\therefore \left( {\frac{{\vec n \cdot \vec b}}{{nb}}} \right)\)
or
Let the given line be \(\frac{{x – {x_1}}}{l} = \frac{{y – {y_1}}}{m} = \frac{{z – {z_1}}}{n}\) and the given plane be \(ax + by + cz + d = 0\). If \(\phi \) is the angle between the given line and the normal to the plane then \(\emptyset = \left[ {\frac{{|al + bm + cn|}}{{\sqrt {{l^2} + {m^2} + {n^2}} \sqrt {{a^2} + {b^2} + {c^2}} }}} \right]\)