Right angle vs. perpendicular vs. orthogonal vs. normal
Right angles, perpendicular lines, orthogonal vectors, and normal vectors all describe things that meet “squarely.” A right angle is an angle of $90^\circ$, so the term describes an angle. Perpendicular describes lines or planes that intersect and form right angles. Orthogonal describes vectors whose inner product is $0$, which extends the idea to vectors that do not meet, such as directions of skew lines, to the zero vector, and to vectors that are not arrows, such as functions. A normal vector is perpendicular to a line, curve, or surface at a point, so the term describes a vector relative to a shape.
Definitions
A right angle is an angle that is a quarter of a full turn, like the corner of a square or of a sheet of paper.
A right angle is an angle that measures
\[90^\circ = \frac{\pi}{2} \text{ radians}.\]Examples
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Two adjacent sides of a rectangle, as segments meeting at a corner
The angle at the corner of a rectangle where two adjacent sides meet is a right angle, because every angle of a rectangle measures $90^\circ$.
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The lines $y = x$ and $y = -x$ in the plane, with direction vectors $(1, 1)$ and $(1, -1)$
The lines $y = x$ and $y = -x$ cross at the origin and form four right angles there.
Non-examples
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The $x$-axis and the line through $(0, 0, 1)$ parallel to the $y$-axis in $\reals^3$
The $x$-axis and the line through $(0, 0, 1)$ parallel to the $y$-axis do not form a right angle, because the lines have no common point at which to form an angle.
Properties
- A right angle is formed by two rays or segments that share an endpoint; the term describes the angle, not the lines.
Two lines are perpendicular if the lines cross and make square corners where they cross.
Two lines in a plane or in space are perpendicular if the lines intersect and form a right angle at their intersection.
Shapes are perpendicular at at a point $P$ if they intersect at $P$ at a right angle.
Examples
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Two adjacent sides of a rectangle, as segments meeting at a corner
Two adjacent sides of a rectangle are perpendicular, because the sides meet at the corner and form a right angle there.
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The lines $y = x$ and $y = -x$ in the plane, with direction vectors $(1, 1)$ and $(1, -1)$
The lines $y = x$ and $y = -x$ are perpendicular, because the lines intersect at the origin and the product of their slopes is $1 \cdot (-1) = -1$.
Non-examples
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The $x$-axis and the line through $(0, 0, 1)$ parallel to the $y$-axis in $\reals^3$
The $x$-axis and the line through $(0, 0, 1)$ parallel to the $y$-axis are not perpendicular, because the lines never intersect: the first line has $z = 0$ and the second line has $z = 1$.
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The vector $(1, 1)$ and the line $x + y = 1$ in the plane, whose direction vector is $(1, -1)$
The vector $(1, 1)$ and the line $x + y = 1$ are not called perpendicular, because “perpendicular” describes two lines or planes that intersect; the normal line through $\left(\frac{1}{2}, \frac{1}{2}\right)$ in the direction $(1, 1)$ is perpendicular to the line $x + y = 1$.
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The functions $\sin$ and $\cos$ on $[-\pi, \pi]$, with the $L^2$ inner product $\ip{f}{g} = \int_{-\pi}^{\pi} f(x) g(x) \, dx$
The functions $\sin$ and $\cos$ are not called perpendicular, because perpendicularity describes lines and planes in geometry, and functions are not lines.
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The vectors $(1, 0)$ and $(0, 0)$ in $\reals^2$
The vectors $(1, 0)$ and $(0, 0)$ are not perpendicular, because the zero vector has no direction, so the vectors form no angle.
Properties
- Two lines in the plane with slopes $m_1$ and $m_2$ are perpendicular exactly when $m_1 m_2 = -1$, or when one line is horizontal and the other is vertical.
- Lines in space whose directions are at right angles but that do not intersect are usually called orthogonal rather than perpendicular.
Two vectors in the plane or in space are orthogonal if the vectors meet at a right angle, which happens exactly when their dot product is $0$.
Consider an inner product space $V$. Two vectors $u, v \in V$ are orthogonal if
\[\ip{u}{v} = 0.\]A list of vectors is orthogonal if every two different vectors in the list are orthogonal.
Examples
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The lines $y = x$ and $y = -x$ in the plane, with direction vectors $(1, 1)$ and $(1, -1)$
The direction vectors $(1, 1)$ and $(1, -1)$ of the lines $y = x$ and $y = -x$ are orthogonal, because their dot product is $1 - 1 = 0$.
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The $x$-axis and the line through $(0, 0, 1)$ parallel to the $y$-axis in $\reals^3$
The $x$-axis and the line through $(0, 0, 1)$ parallel to the $y$-axis have orthogonal direction vectors $(1, 0, 0)$ and $(0, 1, 0)$, so the lines are orthogonal.
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The vectors $(1, 0)$ and $(0, 1)$ in $\reals^2$
The vectors $(1, 0)$ and $(0, 1)$ are orthogonal, because their dot product is $1 \cdot 0 + 0 \cdot 1 = 0$.
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The vector $(1, 1)$ and the line $x + y = 1$ in the plane, whose direction vector is $(1, -1)$
The vector $(1, 1)$ is orthogonal to the direction vector $(1, -1)$ of the line $x + y = 1$, because their dot product is $1 - 1 = 0$.
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The functions $\sin$ and $\cos$ on $[-\pi, \pi]$, with the $L^2$ inner product $\ip{f}{g} = \int_{-\pi}^{\pi} f(x) g(x) \, dx$
The functions $\sin$ and $\cos$ are orthogonal in the $L^2$ inner product on $[-\pi, \pi]$, because $\int_{-\pi}^{\pi} \sin(x) \cos(x) \, dx = 0$, since the integrand is odd.
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The vectors $(1, 0)$ and $(0, 0)$ in $\reals^2$
The vectors $(1, 0)$ and $(0, 0)$ are orthogonal, because their dot product is $1 \cdot 0 + 0 \cdot 0 = 0$.
Non-examples
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The vectors $(1, 0)$ and $(2, 0)$ in $\reals^2$
The vectors $(1, 0)$ and $(2, 0)$ are not orthogonal, because their dot product is $1 \cdot 2 + 0 \cdot 0 = 2$.
Properties
- The zero vector is orthogonal to every vector.
- An orthogonal list of nonzero vectors is linearly independent.
- An orthogonal list of unit vectors is called orthonormal.
Consider a point on a curve. A normal vector at the point is a vector that is perpendicular to the curve at that point. More precisely, a vector $v$ is normal to a curve at $P$ if the line spanned by $v$ from $P$ is perpendicular to the tangent line of the curve at $P$.
Consider a smooth surface $S$ in $\reals^n$ and a point $p \in S$. A vector $n$ is called a normal vector to $S$ at $p$ if $n$ is orthogonal to every tangent vector of $S$ at $p$:
\[\ip{n}{v} = 0 \quad \text{for every tangent vector } v \text{ of } S \text{ at } p.\]Examples
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The vector $(1, 1)$ and the line $x + y = 1$ in the plane, whose direction vector is $(1, -1)$
The vector $(1, 1)$ is a normal vector to the line $x + y = 1$, because $(1, 1)$ is the gradient of $x + y$, and the gradient is perpendicular to each level set.
Properties
- For a level set $\setdef{x \suchthat f(x) = c}$, the gradient $\nabla f(p)$, if nonzero, is a normal vector at $p$.
- The normal line at $p$ is the line through $p$ in the direction of a normal vector.
- “Normal” means “perpendicular” here, and the word is unrelated to the normal distribution and to a normal space.
Examples
| Example | right angle | perpendicular | orthogonal | normal vector | Why |
|---|---|---|---|---|---|
| Two adjacent sides of a rectangle, as segments meeting at a corner | ✓ | ✓ | – | – | |
| The lines $y = x$ and $y = -x$ in the plane, with direction vectors $(1, 1)$ and $(1, -1)$ | ✓ | ✓ | ✓ | – | |
| The $x$-axis and the line through $(0, 0, 1)$ parallel to the $y$-axis in $\reals^3$ | ✗ | ✗ | ✓ | – | The $x$-axis and the line through $(0, 0, 1)$ parallel to the $y$-axis show that orthogonal lines need not intersect, so orthogonal lines need not be perpendicular. |
| The vector $(1, 1)$ and the line $x + y = 1$ in the plane, whose direction vector is $(1, -1)$ | – | ✗ | ✓ | ✓ | The vector $(1, 1)$ and the line $x + y = 1$ show that a normal vector is described relative to a shape: the vector is normal to the line and orthogonal to its direction vector. |
| The vectors $(1, 0)$ and $(0, 0)$ in $\reals^2$ | – | ✗ | ✓ | – | The vectors $(1, 0)$ and $(0, 0)$ show that orthogonality is defined by the inner product alone, so the zero vector is orthogonal to every vector even though the zero vector forms no angle. |
| The functions $\sin$ and $\cos$ on $[-\pi, \pi]$, with the $L^2$ inner product $\ip{f}{g} = \int_{-\pi}^{\pi} f(x) g(x) \, dx$ | – | ✗ | ✓ | – | The functions $\sin$ and $\cos$ show that orthogonality applies to vectors that are not arrows in space, where “perpendicular” and “right angle” have no geometric meaning. |
Notes
- In an inner product space, the angle $\theta$ between nonzero vectors $u$ and $v$ is defined by $\cos \theta = \ip{u}{v} / (\norm{u} \norm{v})$, so nonzero orthogonal vectors form a right angle in this generalized sense.
- Some authors use “perpendicular” and “orthogonal” as synonyms for nonzero vectors.