Angle Between a Line and a Plane

The formula

The angle $\theta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ uses $\sin$ , not $\cos$ , because the angle is measured from the plane itself, which is perpendicular to $\vec{n}$ :

\sin\theta = \dfrac{|\vec{b}\cdot\vec{n}|}{|\vec{b}||\vec{n}|}

The angle between $\vec{b}$ and $\vec{n}$ is $90^\circ - \theta$ , and $\cos(90^\circ - \theta) = \sin\theta$ . That one step is the whole reason the formula switches to sine.

The three angle formulas side by side

Angles in 3D geometry, using direction vectors b and normals n

Angle between	Use	Formula
Two lines	Their directions $\vec{b_1}, \vec{b_2}$	$\cos\theta = \dfrac{\|\vec{b_1}\cdot\vec{b_2}\|}{\|\vec{b_1}\|\|\vec{b_2}\|}$
Two planes	Their normals $\vec{n_1}, \vec{n_2}$	$\cos\theta = \dfrac{\|\vec{n_1}\cdot\vec{n_2}\|}{\|\vec{n_1}\|\|\vec{n_2}\|}$
A line and a plane	Direction $\vec{b}$ and normal $\vec{n}$	$\sin\theta = \dfrac{\|\vec{b}\cdot\vec{n}\|}{\|\vec{b}\|\|\vec{n}\|}$

The absolute value in each one keeps $\theta$ between $0^\circ$ and $90^\circ$ , which is the angle these questions ask for.

Worked example

Find the angle between the line $\dfrac{x+1}{2}=\dfrac{y}{3}=\dfrac{z-3}{6}$ and the plane $10x+2y-11z=3$ .

Solution: The line's direction vector is $\vec{b}=(2,3,6)$ , and the plane's normal is $\vec{n}=(10,2,-11)$ .

\vec{b}\cdot\vec{n} = 2(10)+3(2)+6(-11) = 20+6-66 = -40

|\vec{b}| = \sqrt{4+9+36}=7, \qquad |\vec{n}| = \sqrt{100+4+121}=15

\sin\theta = \dfrac{|-40|}{7 \times 15} = \dfrac{40}{105} = \dfrac{8}{21} \;\Rightarrow\; \theta = \sin^{-1}\dfrac{8}{21}

Worked example: the angle between two planes

Find the angle between the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$ .

Solution: The normals are $\vec{n_1} = (2, 1, -2)$ and $\vec{n_2} = (3, -6, -2)$ :

\vec{n_1}\cdot\vec{n_2} = 6 - 6 + 4 = 4, \qquad |\vec{n_1}| = 3, \qquad |\vec{n_2}| = 7

\cos\theta = \dfrac{4}{21} \;\Rightarrow\; \theta = \cos^{-1}\dfrac{4}{21}

Parallel and perpendicular cases

Line parallel to the plane: $\vec{b}\cdot\vec{n} = 0$ , so $\sin\theta = 0$ . The line either lies in the plane or never meets it.
Line perpendicular to the plane: $\vec{b}$ is a multiple of $\vec{n}$ , so $\theta = 90^\circ$ .

Example: is the line $\vec{r} = \hat{i} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ parallel to the plane $x + 3y + z = 4$ ?

Solution: $\vec{b}\cdot\vec{n} = 2(1) + (-1)(3) + 1(1) = 0$ , so yes. To see whether it lies in the plane, test its point $(1, 0, 0)$ : $1 + 0 + 0 = 1 \neq 4$ . So the line is parallel to the plane and never meets it.

Common mistakes

Using cos for a line and a plane. That gives the angle between the line and the normal, which is $90^\circ - \theta$ .
Reading the direction from the wrong numbers. In $\dfrac{x+1}{2}=\dfrac{y}{3}=\dfrac{z-3}{6}$ the direction is the denominators $(2, 3, 6)$ . The numerators only give a point.
Dropping the absolute value. Without it, $\vec{b}\cdot\vec{n} = -40$ would give a negative sine and an angle outside the range the question wants.
Calling $\vec{b}\cdot\vec{n} = 0$ perpendicular. For a line and a plane, a zero dot product means parallel. It's perpendicular when $\vec{b}$ and $\vec{n}$ point the same way.

Where this comes up in exams

The JEE Advanced 2026 syllabus lists the angle between two lines, between two planes, and between a line and a plane. The JEE Main 2026 syllabus and the 2025-26 CBSE Class XII syllabus list only the angle between two lines.

Checked against JEE (Advanced) 2026 syllabus, JEE (Main) 2026 syllabus (NTA) and CBSE Mathematics curriculum, Classes XI-XII (2025-26). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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