To find a vector orthogonal (or perpendicular) to a given vector, you need to find a vector whose dot product with the original vector is zero. As stated in the reference, two vectors x and y in R^n are orthogonal if x ยท y = 0.
Understanding Orthogonality
Orthogonality is a fundamental concept in linear algebra and geometry. It describes vectors that meet at a right angle (90 degrees).
The Dot Product Rule
The key to finding an orthogonal vector lies in the dot product. The dot product of two vectors $x = [x_1, x_2, ..., x_n]$ and $y = [y_1, y_2, ..., y_n]$ in $R^n$ is calculated as:
$x \cdot y = x_1y_1 + x_2y_2 + ... + x_ny_n$
For vectors to be orthogonal, this sum must equal zero: $x_1y_1 + x_2y_2 + ... + x_ny_n = 0$.
Methods to Find an Orthogonal Vector
Finding an orthogonal vector involves setting up the dot product equation and solving for the components of the unknown vector. Since there are typically multiple solutions (in fact, an infinite number in $R^n$ for $n > 1$, unless the original vector is the zero vector), you usually choose one or more components of the orthogonal vector and solve for the others.
Note: As the reference mentions, the zero vector is orthogonal to every vector in R^n because $0 \cdot x = 0$ for any vector x. While this is technically a valid orthogonal vector, usually when asked to "find an orthogonal vector," people are looking for non-zero examples.
Example in 2D ($R^2$)
Let's say your given vector is $v = [a, b]$, where $a$ and $b$ are not both zero. You want to find a vector $u = [x, y]$ such that $v \cdot u = 0$.
The dot product equation is:
$a \cdot x + b \cdot y = 0$
To find a solution, you can pick a value for $x$ or $y$ and solve for the other.
For instance, if you let $x = b$ and $y = -a$:
$a(b) + b(-a) = ab - ba = 0$. This works!
So, a simple orthogonal vector to $[a, b]$ is $[b, -a]$. Another is $[-b, a]$. Any scalar multiple of these will also be orthogonal (e.g., $k[b, -a]$ for any scalar $k$).
| Given Vector $[a, b]$ | Orthogonal Vector $[b, -a]$ |
|---|---|
| $[2, 3]$ | $[3, -2]$ |
| $[-1, 5]$ | $[5, 1]$ |
| $[4, 0]$ | $[0, -4]$ (or $[0, 1]$) |
Example in 3D ($R^3$)
Let your given vector be $v = [a, b, c]$, where $a, b, c$ are not all zero. You want to find $u = [x, y, z]$ such that $v \cdot u = 0$.
The dot product equation is:
$ax + by + cz = 0$
This is a single equation with three unknowns ($x, y, z$). This means there are infinitely many solutions (they form a plane through the origin). To find a specific solution, you can choose values for two variables and solve for the third, or use a more systematic method like the cross product (though the cross product only works in $R^3$).
Method 1: Choose Components
- Set one component of the orthogonal vector to a non-zero value and one to zero.
- Solve for the third.
- If $v = [1, 2, 3]$, find $u = [x, y, z]$ such that $1x + 2y + 3z = 0$.
- Let $x = 1$, $y = 1$. Then $1(1) + 2(1) + 3z = 0 \implies 3 + 3z = 0 \implies z = -1$. So $u_1 = [1, 1, -1]$ is orthogonal: $1(1) + 2(1) + 3(-1) = 1 + 2 - 3 = 0$.
- Let $x = 0$. Then $2y + 3z = 0$. We can pick $y=3, z=-2$. So $u_2 = [0, 3, -2]$ is orthogonal: $1(0) + 2(3) + 3(-2) = 0 + 6 - 6 = 0$.
- Let $y = 0$. Then $ax + cz = 0$. Pick $x=c, z=-a$. If $v=[1,2,3]$, pick $x=3, z=-1$. So $u_3 = [3, 0, -1]$ is orthogonal: $1(3) + 2(0) + 3(-1) = 3 + 0 - 3 = 0$.
Method 2: Using the Cross Product (Specific to R^3)
The cross product of two vectors $v$ and $w$ in $R^3$, denoted $v \times w$, results in a vector that is orthogonal to both $v$ and $w$. If you only have one vector $v$, you can take the cross product of $v$ with any non-parallel vector $w$ to get a vector orthogonal to $v$. A simple choice for $w$ could be a standard basis vector like $[1, 0, 0]$, $[0, 1, 0]$, or $[0, 0, 1]$, provided $v$ is not parallel to it.
Let $v = [a, b, c]$.
If $v$ is not parallel to $[1, 0, 0]$ (i.e., $b \neq 0$ or $c \neq 0$), take $w = [1, 0, 0]$.
$v \times w = [a, b, c] \times [1, 0, 0] = [(b)(0) - (c)(0), (c)(1) - (a)(0), (a)(0) - (b)(1)] = [0, c, -b]$.
Check the dot product: $[a, b, c] \cdot [0, c, -b] = a(0) + b(c) + c(-b) = 0 + bc - cb = 0$. This works!
If $v$ is parallel to $[1, 0, 0]$ (i.e., $v = [a, 0, 0]$), you would need to pick a different vector for $w$, like $[0, 1, 0]$.
$v \times [0, 1, 0] = [a, 0, 0] \times [0, 1, 0] = [(0)(0) - (0)(1), (0)(0) - (a)(0), (a)(1) - (0)(0)] = [0, 0, a]$.
Check the dot product: $[a, 0, 0] \cdot [0, 0, a] = a(0) + 0(0) + 0(a) = 0$. This also works!
In general, for $v = [a, b, c]$, vectors orthogonal to $v$ satisfy $ax + by + cz = 0$. Common simple non-zero solutions are $[-b, a, 0]$ (if $a$ or $b$ is non-zero), $[-c, 0, a]$ (if $a$ or $c$ is non-zero), and $[0, -c, b]$ (if $b$ or $c$ is non-zero).
Summary of Finding Orthogonal Vectors
- Given vector $v$.
- Define an unknown vector $u$ with components.
- Set up the dot product equation $v \cdot u = 0$.
- Solve the resulting linear equation for the components of $u$.
- Choose specific values for some components (often setting some to 0 or 1) to find a particular solution.
- Remember that the zero vector is always a solution, and usually, there are infinitely many non-zero solutions forming a subspace orthogonal to the original vector.
